 So we've been looking at the rules of implication and the rules of implication in an important sense take what we already know and because of their relationships between the say the different propositions between what we already know because of their relationships they tell us something new. So from a conditional and the assertion to the antecedent we can infer the consequence that's what opponents from any assertion we can infer a disjunction a disjunction that contains that assertion. So we take what we already know and then we infer something new with the rules of implication the rule equivalence rules are slightly different. So like I said equivalence rules a little bit different implication rules of implication we're able to take what we already knew and infer something new. So they work pretty much just one way right from a conditional and the assertion of the antecedent we can infer the consequence. We can't just take any proposition we'd like so that's what opponents work in one way but we can't go the reverse right it can't just take any proposition we like and infer you know a conditional and the assertion antecedent with a disjunctive introduction we could take a single proposition and infer a disjunction that contains that proposition but it just doesn't work that way it doesn't work the other way right I can't take a disjunction and infer just one of the disjuncts. Rules of application give us something new with what we already knew equivalence rules don't give us anything new the equivalence rules don't really tell us anything new because the formulas that we're able to switch out have the exact same truth values so in other words I mean if you go back to the truth tables no matter the truth assignments you have for p and q the truth value the compound propositions will be exactly the same. Since they have the exact same truth values since there's really nothing new that we know or we're learning about these formula we can simply swap them out. They can they go in both directions so for instance one of the rules we're going to see is double negation this the simplest equivalence rule so it just take any proposition and we've seen a negation of a proposition well if you negate the negation right so minus sign minus sign proposition that proposition that double negation has the exact same truth value as the proposition so you can just simply swap it out and you can swap it out swap it out in the middle of any formula right it could be the proposition that you swap out or the part that you swap out could just be part of it a part of a you know a larger complex formula or it could be the whole thing so uh the equivalence rules are going to be different than the rules of implication rules of implication go only one way with the rules of implication can't just swap things out right can't swap out parts but the equivalence rules they go both directions right so with any double negation you can just you can just swap out for the for the single proposition and for any single proposition you can swap out the double negation no problem uh so it goes both ways and you do within parts of a larger complex formula that's just fine so with the rules of equivalence we're able to swap out formula propositions with their logical equivalent formulas right so one is as good as another in terms of truth value and sometimes the results might be surprising but in terms of truth value we're not learning anything new uh that's this is different from the rules of implication because we're learning something new in terms of you know something in a sense something new with truth values with uh these further propositions now since the rules of equivalence are different especially meaning from the rules of implication they are going to be different in their use and application within the proofs so with that in mind let's take a look at a few problems you know look at a few problems along the way with each of the rules and see how that's going to look so these are the rules of equivalence that we're going to be using uh in in this course now there's a handful of them and you're going to want to try and keep track of them so except my nose or something like that I wouldn't try and do this from memory if I were you well the first one I'm going to look at is double negation I already mentioned that as an example you know double negation allows us to you know swap out any proposition with its negated negation so I'm outside here's a proposition I am outside here's the negation I am not outside here's the double negation it is false that I am not outside okay uh being outside is kind of one of the things you are you aren't so uh double negation allows us to swap out any proposition with this negated uh negation and you know this finally lets us do something that we probably thought was really obvious up to this point uh with modus tollens so let's let's take a look at this problem here so we have uh we have a uh conditional it's a conditional negations and and we have the uh assertion of the consequent so you have if not q then not p and we have p and we're from this replacement for q so how do we do this right because right now we because so far we're just modus tollens we weren't able to do this modus tollens allows us to take a proposition with the negation of its consequent and in front of the negation of its seed but we don't have a negation of its consequent here right the negation of the consequent would be not not p that would be not not p well so we have our first two premises all right well with double negation we are now able to just swap out that p with not not p right so we can put that in the that third line there and since we now have the negation of the consequent I mean we can infer the negation of the intercedent which is not not q now since what we're looking for is q right we're not our work isn't done yet well fortunately our work is really simple now we just swap out q for the not not q now double negation is probably the easiest of the equivalence rules to deal with um so let's move on to the next rule well it's not the last time we see double negation you know either i'm walking over a bridge or i'm walking over a bridge also i am walking over a bridge and i'm walking over a bridge might say i'm just repeating myself well that's kind of the point of this next rule redundancy from any disjunction where both disjuncts are the same proposition we can just infer that proposition and vice versa you know from any proposition we can infer disjunction of that proposition um same thing with conjunction all right from any conjunction where each proposition or both propositions are the same conjunction conjunct we can infer just that conjunction so probably one of the ways that this is most or readily applicable is when we have something like you know like this problem here where we have dilemma so far every problem with dilemma that i've given you uh except if i remember correctly you know it's just resulted in another disjunction and that's not bad but what happens when each disjunction where both excuse me both disjunctions are the same so you know we have a p then q and p then r then q then r we need to infer r up to this point we haven't been able to do that right we have to infer the disjunction r or r well you know let's just quickly follow the steps through the dilemma because we've seen this before remember we have our three uh premises well you assume for the sake of disjunction that's that ad means you assume for the sake of disjunction the left hand disjunct and from that uh we use the conditional if p then r to infer r you know two and four using modus ponens uh then so that that half of the job is done let's assume the right disjunct and the right disjunct is q and again using that other conditional of q then r so you know three and six using modus ponens we've got the other r right so then we have now we have our disjunction r or r right r or r and that's from the first premise where we get the disjunction the fourth line where we assume the left hand disjunct the fifth line where we get the you know what we wanted to conclude from that left hand disjunct the sixth line where we assume the right hand disjunct and the seventh line where we infer what we need to infer for this the right hand disjunct and then we cite that with the lemma right so we've got our conclusion well we're still looking for r well now we can use the equivalence rule right and the equivalent rule comes in and we're able to just reduce that down from r or r because you know the way it is at least one of these is true right well at least one of these is true then you know both are whether we get to infer r so with redundancy we're able to take that disjunction down to just the single proposition now you know so far we've just been using these equivalent rules to kind of clean up the conclusion to get what we want it's going to get more involved right these are just introductory examples uh to show you a little bit how they work it is going to get more involved the further along we go and I think we're going to see that uh more and more we get to the more complicated equivalence rules the next one's commutation commutation it looks obvious so obviously we don't need to bother mentioning it but there's reasons for it so with commutation whenever we have a disjunction or a conjunction right we can swap the positions of each one so if we have p or q we can infer q or p and vice versa if we have p and q we can infer q and p or vice versa so commutation allows us to swap they say notice conditional is not in this rule right we can't just swap positions with a conditional and get the same thing so don't confuse commutation with conditions so let's take a look at a look at a problem with this okay so we have not not p hmm I don't know if we can use double negation there right we have not not p and then if q then q or r excuse me and then the second premise if either q or p then r right and so we need to infer r well already you're probably probably gonna need to have more than one equivalence rule working here that that not not p is kind of a glaring thing right so so as with all these problems take a look at r or excuse me take a look at the conclusion that's r and we want to be able to pull out that conclusion from the premises well where is it well is the consequent of a conditional in that second premise okay cool right so since it's the consequent of a conditional we're probably going to use modus ponens to get it so what's the antecedent of that conditional it's q or p we look at the rest of the premises it's not there but we do have that not not p we have that not not p and we also have the inference rule disjunction introduction so that's probably we're going to get that antecedent well let's give it a shot so first we'll take p excuse me a technique a not not p and with the equivalence rule of double negation all right for that first line we'll infer p and again it could go both directions but we're just using one direction this time right well so we got p and we still need you know the p or q well let's use double negation excuse me let's use disjunction introduction and infer that p or q okay great well p or q was technically not uh you're not exactly the same as that q or p i mean it isn't it isn't right uh it's the exact same in truth value well since it's the exact same in truth value let's swap out swap the positions of of the p and q there and use q or p with our equivalence rule so say well it means the same thing well of course it means the same thing that's why we have the equivalence rule i mean that's the point of these equivalence rules to be able to state you know as explicitly as possible this is the same thing and it works out in the truth tables uh they have the exact same truth value so it's fine right it's fine and probably also and again you have this rule specifically for disjunctions the conditionals to emphasize the fact that we can't use these with excuse me have this rule for disjunctions and conjunctions disjunctions and conjunctions but not with conditionals right we can't use the this rule as it stands with a conditional so now we have our our q or p well then it's simply a matter of uh modus ponens at this point with line two and line five to infer r uh so you know just maybe emphasize this fact we can use this more than one equivalence rule and or proof and as we'll see probably later on you can use uh the equivalence from more than once in our proof right all right so that's uh that example uh that's that's commutation we could switch positions of the propositions for disjunctions and conjunctions so it's next rule association lets us deal with three disjunctions or three conjunctions conjunctions excuse me at a time so so far we we really haven't been doing this our operator our logical operators only deal with two propositions at a time uh but if we've got three disjuncts association will help us handle these three disjuncts if we've got three conjuncts association will help us handle them we have to do it a step at a time but you know it'll help us handle these even though our operators only deal with two propositions at a time so uh let's take a look at how this how this works so let's take a look at this problem so we've got p you know got p or and then the further disjunction q or r and we've got the conditional if uh p or q then s and the third and the second conditional if r then s and what we want to infer is s so look at that conclusion s where do we find it we find it as the consequent of two conditionals and where do we find the antecedents the antecedents are in that first premise just not presented the right way right so we have and that first premise we have p as one disjunct and then q or r as the other disjunct and it'd be great to be able to use delimit this point to pull out that s but i don't have p or q as a disjunction i don't have r as a disjunction but wait there's association so uh let's first take that that first premise association right and uh this allows us to take p or and the disjunction q or r and infer p or q disjoint to r and it works both directions just keep that in mind it can go both directions so we've got uh now we've got the right disjunction to use dilemma so we assume p or q for the sake of dilemma and using that assumption and premise two we're able to infer s that's the left-handed part of the disjunction then we assume the right-handed part of the disjunction assume r for the sake of disjunction we're able to infer s using that conditional in line three so we got our s or s now notice the citation the citation is not the initial disjunction or premise one i mean it's not uh it's premise four now four is equivalent to one but the disjunction we're using for dilemma is that premise is that line four not line one so it's line four line five is where we assumed uh the left-handed disjunct line six is where we inferred our s what we're looking for from that left-handed disjunct line seven is where we assume the right-handed disjunct line eight is where we inferred our inference from that disjunction right so that's how we're able to infer s or s and then now using redundancy which we already learned we're able to just infer s so that's one way that association could work let's look at another problem that one dealt with disjunctions this one does with conjunctions so here we have p conjoined to the conjunction q and r and then we simply have the conjunction p and q in for you know part of a conditional is the antecedent and s is the consequent and we need to infer s this one's pretty straightforward it's easy to find s and our premises and we got p and q as the antecedent but p and q at least not explicitly is there but we do have that conjunction so we take that conjunction and again using an association we move right the parentheses over and now p and q the conjunction p and q is conjoined to r and so we separate out p and q using conjunction elimination and from that in the conditional and premise two we're able to infer s and two pretty straightforward uses of association but wait there's more let's take a look at the third problem so now we have just a single premise p and q and from this we're supposed to infer p or r disjoined to a further proposition q or r right p or p or r either p or r or either q or r it's a little cumbersome to say in english which is probably why you know having this notation is a good thing okay so how are we going to do this well looking at the conclusion you might notice that r is to be found nowhere in the premises we got p or q but we don't have r anywhere in the premises well that that gets a little strange right uh how are we supposed to uh deal with that then um if you know if we got something out of nowhere right there's only two rules that'll let us deal with something out of nowhere well disjunction introduction is one of them and uh conditional proof is the other right conditional proof we're able to assume the antecedent to infer the consequent and that allows us to infer the conditional well disjunction introduction allows us to refer disjunction from any given premise okay so since it's just purely out of nowhere and we're not dealing with the condition well then yeah we're probably dealing with disjunction introduction so let's take the initial prop premise p or q and then what do we do well we from that we infer just uh p the disjunction p or q or r right that's our inference from disjunction introduction well don't we need two r's well yes we do i'm glad you asked that so we need two r's well then we're able to use that r and swap out the single r for a disjunction of r or r using redundancy so this is the first time we've got like that little swap out just within uh just within part of a complex proposition this works for equivalence rules this does not work for rules of implication don't try to swap things out using rules of implication only the equivalence rules right only the equivalence rules this does not so with the equivalence rules we can swap out uh these parts this does not work for the rules of implication we can't swap it out with the rules of implication only with the equivalence rules all right so now what do we do well we've got a kind of complex set of parentheses here so i'm just going to show you the next move it's that line four now this might look confusing but this works using association where uh you know the right most right-handed r there is treated as one of the disjuncts and the left mode you know so then we got that complex besides the p or q or r that's treat the p or q is treated as the left most disjunct well we can just you know double the parentheses over around that right so we've got the p we've got the p or q as one disjunct disjunct to r and then that whole disjunction is also disjoint to r well now we use uh good old association again and this time we move the parentheses around q or r so we got p as one disjunct then we've got the q or r right next to it we saw that that last r dangling out sure wish we can move it on over to the other side and then that way be really easy to use association but wait we can after all we have commutation commutation allows us to switch positions the destructions so we switch the r with the other you know complex problems with the p of the q or the r right so we switch that around in line six and then again using association we're able to move those parentheses out uh to uh excuse me new parentheses out to conjoin r to p right and then using commutation just one more time switch positions of p or r so now we have p or r could join to q or r that was a lot of equivalence rules but sometimes that's the way it works let's look at one more problem this time this should look kind of familiar but instead of the disjunctions this time now we got the conjunctions right now we got the conjunctions so this problem looks a lot like the disjunctions except now we got conjunctions right so you can probably figure out what we're going to do so we got the p and q can join to r and we want to infer p and q and p and r excuse me p and r and q and r excuse me p and r and q and r okay so how are we going to go about doing this well we got the initial premise p and q could join to r um but we sure need two r's well think about this if you ever need two instances of the same proposition this is where redundancy comes in handy right we just simply doubled it up right um so now so in line two what we're able to do is we get get the conjunction of r and r from the single and so and again since it's it's one part using the equivalence rule we can swap it out we could do that with equivalence rules we can't do it we could do that with rules of implication okay so now again using association we've got the premises around p and q and r and that last and r is on the outside and using association again instead of p and q can join can join to r now we got p can join to q and r we still got the r hanging out the other side well we need that r back over on the other side at this at the front well to get that in the front then we use commutation so move that on over uh in line uh line five line six we need the parentheses around r and p so with association again we can have the parentheses around r and p but we need p and r to switch places now now we can use that uh do that using commutation again so now we got commutation again so uh this time you know here here's here's something fun not a single inference rule is used in this proof so this is what i was saying earlier uh we might get surprised by some of the results with equivalence rules but technically speaking that's the same exact truth values if you took these two propositions or the premise and the conclusion and put them on the truth table you'll have exactly the same truth values and the equivalence rules reflect that and allow us to swap out these premises so that's so so far we got double negation redundancy commutation and association i would start to enter some of the really fun stuff so so far with uh commutation and association we're dealing only with uh one kind of logical operator right it might be several instances of it but only kind of one one kind of logical operator all either all disjunctions or all or all conjunctions okay this next rule distribution deals deals with a mix of two operators and that's with a mix of conjunctions and disjunctions sorry so the first way to we're going to think about this is when we got a conjunction where one of the conjuncts is a disjunction and when that's the case we can distribute so the right-handed conjunct is a disjunction we can distribute the left-handed disjunct excuse me the left-handed conjunct to the two disjunctions so if i got p can join to the disjunction q or r from that's equivalent to a disjunction and the disjunction is p and and let's say the disjunction is q or r that's equivalent to a disjunction p and q or p and r and so the idea is at least q or r or true is true and we already know that p is true so we can refer to disjunction now either p a q is true or r is true well we can could join q or r in this case or could you can join p or r in that case right so this also works the other direction right from a disjunction where each disjunct is a conjunction and each and in each of the conjunctions you have the same premise we can pull out um that pull out that conjunct so it works in both directions the other way distribution works is when we're dealing with a disjunction so we had the left handed disjunct and it's just joined to a conjunction so we got p or q and r right well we can right distribute that disjunct over the conjuncts so you got p and q or uh p and r so let's see this with a homework problem so we've got two premises here first premise is the disjunction p disjoint to q and r g i wonder if we're going to use distribution there second premise is p or q and that that's the that's the antecedent of a conditional with s as the consequence and the conclusion is s okay where we already identified right s is the conclusion we need we already identified where s is in the premises so it's as the consequence of that conditional to get that consequent out we need the antecedent p or q well p or q is not present now one way it's gonna be honest one way to do this right well yeah so we think maybe that one way to do this is to uh do a dilemma right but we don't have any conditionals that allow us to infer that q and r to get to s right we don't have any of that so we need to use distribution in this case right so we've got line three it's where we use distribution so from that disjunction we're able to refer a conjunction p or q could join to p or r and from that conjunction we could pull out p or q using conjunction elimination and then it's a simple matter of that premise two with the conditional their premise two to infer s using modus ponens all right so that was pretty straightforward let's try another problem all right it's a little bit longer this time this time we've got a conjunction with p and it's conjoined to q or r and now we have two conditionals p and q implies s and then p and r implies also implies s and the conclusion is s all right well where's the conclusion that conclusion s is as a consequent of two conditionals now this this can be a clue right if there's two conditionals there maybe may probably you know the good chance that you could be using dilemma at this point right but we don't have a disjunction of p and q or p or r p p and r excuse me p and q or p and r well we kind of do with that first premise right with that first premise we could distribute that p across q and r and we get a nice big disjunction right so that's what we have in line four p and q is just joined to p and r all right great well now it's just a simple matter of using dilemma so we assume for the sake of dilemma p and q and from that assumption and that second premise there we're able to infer s so that job's done let's get the right hand let's use the right handed disjunct so we assume p and r for the sake of dilemma on the right hand side and from that using that third premise we're uh yeah third premise able to infer s again and since we assumed s from both of these disjuncts using using assuming that they're true right well able to infer the disjunction s or s well we just need the s well gratefully we have redundancy we could take that s or s and reduce it down to r so we used um two equivalence rules there all right so that's uh uh distribution and so far we've been dealing just with negations uh conjunctions and disjunctions let's take a look at conditionals okay now we're getting to an equivalence rule that deals with the conditionals and a conjunction and that is uh uh exportation exportation might look a little confusing so let's think of it this way let's start with the left hand side so we've got a conditional where the antecedent is a conjunction now if you remember right our conjunction is true just in case both conjuncts are true so in order for that antecedent to infer the consequent now we need both conjuncts well it doesn't matter whether both conjuncts are true at ones or true in sequence right this sort of thing doesn't really happen in logic it's just whether they're true or false so with the conjunction excuse me with the conditional where the antecedent is a conjunction we can pull out the left hand side conjunct as its own antecedent and then the consequent is a conditional with the other conjunct as sufficient for r so we already so the idea is well we already have p we know p is true well now we gotta infer q in order to we gotta have q in order to infer r right so that's that's one way and we one one direction the other direction right we got a conditional we're just p and the uh consequent is another conditional q or r well that's equivalent if we have both p and q right if we have both p and q it's not hard to infer r right after that so that that's exportation um hopefully that it's pretty obvious what's happening um if not you can sit down with the truth table and see that these are in fact uh equivalent these are in fact equivalents let's look at some homework problems so we've got two premises uh p or you know a p excuse me p implies q or r and we've got a pretty long conjunction p and q and s right and what we need is r okay well we need r and where is r in our premises well it's a consequent and a conditional that is itself a consequent so we don't need the conditional we need that whole consequence we need the consequence of that condition well in order to get that right we're going to need both p and q okay so that that brings us to the second premise we got p and q right there all right well let's go and pull out that p and q now yeah we could do this using conjunction elimination in a couple of different steps okay you you can but yeah hey let's use equivalence or just along over here so let's use commutation on on that line two right line three we got commutation and we got p and q is one conjunct and then s is the other now let's use conjunction elimination to get that p and q out so now we get we've got the p and q out great now again we could just you know pull it apart and then use modus ponens twice to to get the r but hey since we've got the equivalence rules let's let's make it a little easier on ourselves and let's short down our work a little bit so let's use that line one on line one let's use exportation and with exportation we're able to put p and q as the antecedent and then s is the consequent excuse me r is the consequent right r is the consequent and now we have the really easy job of simply inferring r using that conditional line five with the with the antecedent that conjunct the antecedent line four and now we're able to infer r using lines four and five okay so that you know one way to think about it shortens down a little bit of work here all right well let's try another one right and this works a little bit differently this time we're not trying to get r out we're just trying to get the the consequence that the q or r out right so that's our that's our problem here q or r well again look at the conclusion q or r and that's not contained as a in the premises is not contained as a conditional end of itself but we got p and q infers r okay well it's sufficient for r well let's use exportation let's use exportation on that first premise then we got p is sufficient for q or r and we do you know we got p in line two so from that using modus palanins we're able to infer if q than r another little handy trick keep mine here maybe commutation is going to come up so maybe the uh conditional isn't going to be p and q but it's going to be q and p right so maybe these commutations switch it around that could happen right just just keep that in mind right whenever you got conjunctions or disjunctions that's a handy occasion to use commutation all right or at least that's in my mind is like I don't want to make any commutation there so just keep that in mind all right uh so that's one and that's one equivalence rule using uh conditionals let's look at another one so exportation dealt with conditionals and conjunctions this next one just deals with uh conditionals now you remember with commutation all right we could just simply swap the positions of disjunctions or disjuncts and conjuncts right if you have a disjunction you could swap the positions of the disjuncts you have a contract you could swap the position of the contracts and I said that you know let's not do this with conditions they can't do this with conditions and you can't right at least not simply we can do a little swapping with conditionals but the truth values are not going to remain the same so if we have a conditional right I could swap the interesting for the consequence but I have to negate them right I have to negate them if I've got if p then q I can swap their positions but if I do so then I got to have not q implying not p right if I've got a conditional of negations I can swap them right and in swapping them I just have the the simple assertions I don't have the negations say this is called contraposition contraposition we're able to swap the entity for the consequent but negations are involved right if you're swapping assertions the swap results in negations if you're swapping negations the result the result is assertions okay so let's look at a problem with this so we've got p and q is sufficient for r and what we want to infer is sorry and we got p and we got p but what we want to infer is if not r then not q so this should look a little familiar right because we just dealt with something like this with the last problem so we got that p and q and already that p and q that should send up a little red flag I can use an exportation on that and look at the conclusion I got if not r then not q well that conditional is not anywhere in there but we've got you know r as a consequent in the first premise so probably should be already be on alert probably going to use exportation here and what do you know that's what we do line three but export that p out you get that p out of there well then using lines two and uh two and three excuse me yeah using lines two and three modus ponens we're not able to infer if q then r okay great but I need not r if not r then not q all right well this is where exportation comes in handy right now we can infer that if not r then not q using exportation so it's a little bit of a simple example using exportation it can get more complicated pretty fast we'll see that okay we've already seen an equivalence rule uh involving a conditional now conditionals and conjunctions we also have equivalence rule involving disjunctions and conditionals although just to be clear it's not you know distributing a disjunction over the end to see the consequence that's not it's not like that rather it's that every conditional is equivalent to a disjunction so we have a conditional if p then q and this is equivalent to a disjunction of not p or q now this might look strange but remember our truth conditions for a conditional a condition is true just in case either the antecedent is false or the consequence is true well that is contained in that disjunction similarly we have that with a disjunction if you have a disjunction the negation the negation the antecedent that excuse me and the negation of the of the left hand disjunct that is equivalent to a conditional right where that left-handed disjunct the negation in the left-hand disjunct is now an assertion and that uh implies q and this is just that you know this is basically disjunctive syllogism right disjunctive syllogism if you have the negation of the left-handed disjunct the right-handed disjunct follows okay so now you know keep in mind commutation this should be in the background if you have at least one negated conjunct that's a commutation away being turned into a conditional using uh material implication so let's let's look at a homework problem with this okay so we just simply have p right the antecedent and now we imply uh now from that we infer if q then p i mean this looks kind of weird uh but it's really not as strange as you might think um so it how do we do this well we have our you know have our p right that's our premise well from this we can infer uh using disjunction introduction we can infer p or not q right remember a disjunction introduction we can have a disjunction with anything that we like and that's fine right well now we have not q as our disjunction there's a rabbit i don't know y'all can't hear it on this microphone but there's a rabbit over there i can hear him moving around uh or some little critter okay so uh so we have p disjoint and not q well let's use commutation right uh conditional just a commutation away commutation now we have not q or p not q disjoint to p so now we can use our material implication on that on that disjunction to get if q then p okay let's take a look at another problem all right so here's here's another version of this right so now we have p by itself and this implies if not p then not q so this should kind of already make sense right we've already seen a version of this with the previous problem because now we just pretty much just follow those steps right we've already seen how to get if q then if q then p well remember good old contraposition right contraposition allows us to swap places with the antecedent the consequent but then have the negation in there so we got our line one right that's our assertion from this this should look real familiar right use disjunction introduction to infer a p or not q commutation switch places now we got not q or p from this using material implication line three with material implication if q then p and now we apply contraposition to get if not p then not q right so yeah this is something to keep in mind if you've got a conditional of negations contraposition might be in your future or if you've got just simply if you've got so you know works but both ways right if you got a conditional and the conclusion contains those same parts of the conditional but the negated in the different places yeah contraposition is probably in your future if you've got a conditional of negations and in the conclusion you've got the conditional they're swapped but no longer negated yeah contraposition is probably in your future probably in your future okay so that was that was kind of a variations of the second process kind of variation the first one okay so now we just have p or q p and q let's look at this another problem we got p and q and now we got a conjunction of just of conditionals hmm that's strange well think about it every conditional is equivalent to a disjunction so you can really look at that conclusion it's a conjunction of conditionals but that's also equivalent to a conjunction of disjunctions haven't we seen something before that allows us to distribute a disjunction over conjuncts there it is right there it is okay so uh you know so we're we've got distribution in mind right we're pretty sure we're gonna use distribution here all right well let's take our first premise we got p and q well we we still need that r right we're pulling r out of nowhere well if you're pulling r out of nowhere chances are you're using disjunction introduction right okay so we've got uh that p and q disjoint to not r all right great and we say not r you know just keep that in mind if you're gonna have what you're pulling out of the air as an assertion you you need to if you're using disjunction introduction you need to introduce a negation of that same thing okay so we've got we've added on our disjunction okay well that not r's in the wrong place we need it on the other side well that's what commutation comes in play so commutation allows us to move that negated disjunct to the other side and that's what we do in line three now we can distribute right so we distribute that r across the conjuncts and what we get is a conjunction of disjunctions not r or p can join to not r or q right okay so the next step you might think what would you use material implication on both no we're not going to do that if you can use an equivalence rule use it a one one part at a time right so we'll use the equivalence we use material implication on the left-handed conjunct it's it's best to work from left to right when applicable work left to right when applicable so we use material implication on that left-handed conjunct and now we got the conditional r than p can join to the disjunction not r or q all right well we need the other conditional so now we use material implication again to get that not to turn that disjunction into a conditional so now we have our conclusion if r than p can join to if r than q easy peasy all right let's look at another homework problem okay this looks interesting now we got just a disjunction right and the disjunction we we're turning that disjunction into a disjunction of conditions now that r is still coming so you look at the the conclusion right r than if r than p is disjoint to if r than q well again that r is coming in now we're so we're probably still using disjunction introduction except this time we're not distributing conjunctions over disjunctions or disjunctions over conjunctions right it's all disjunctions so what are we going to do right we can't because we can't distribute disjunctions over disjunctions at least not with a rule but we can do it with the other equivalence rule so we have remember association that's how we're able to get disjunctions you know moved around with their parentheses and we need two r's then well if we need two r's that's where redundancy comes in right now now you start to see where some of these problems can get complicated pretty fast so first things first right so we have our p or q well let's go ahead and disjoint that to not r that's going to be our big thing right disjoints to not r okay second so next after that use redundancy we had not r before well let's just join not r to not r we could do that okay so then by association we're moving around the parentheses by association moving those parentheses around to to disjoint p or q to to one of the not r's and the other not r's just just joined all off on the side then by commutation we move those around move that that not r on the left hand side of that p or q and by association again we get the parentheses around one not r to the q right then by uh this time we just got a simple association move over right just moving the parentheses over uh around the q and then not r and one little bit of commutation switched in the place between the r and q the not r and the and the q right in line nine now this is where we apply material implication on each part kind of one question the same thing we like for the computation the association can't just use one line of association to just move it all the way no you got to do it one step at a time right you do it one step at a time so uh material implication the left-handed disjunct followed by the material implication on the left on the right-handed disjunct and now we got our conclusion i want to pay attention to these examples this shows you kind of the ninja moves we could take with association and commutation all right let's take a look at one more problem this time we're starting with a conditional and what we're inferring is a not only uh not only is the consequent now in the antecedent and negated hand hand hand contraposition but it's conjoined to another negation well how does that happen how do we get a how do we start with the conditional and get a conjoin a conjunction as the antecedent that looks interesting well one step at a time right so first of all we need the antecedent the consequent right and in the in the right order and negated so let's go and use material implication like on the p and q so we got q and p if not if if not q then not p right so where we're going to get that r well remember if we're pulling something out of the air disjunction introduction is a nice choice so let's just join that to r now we're destroying it to r this time because we need the negation of r right we need the negation of r the antecedent if you need a negation in the antecedent using disjunction introduction to get this if you need a negation then you infer then you uh introduce the assertion and maybe maybe maybe double negation is in your future okay so we got r there all right so now we uh but it's in the wrong place so let's use commutation swap the r on the left hand to the left hand side it's in the right place but it's the wrong uh it's the wrong value it's not a negation well let's go ahead and double negate that bad boy double negate that r and again you're just look we can apply the equivalence to just in one part of the whole thing of the complex proposition not just uh not not the entire proposition okay so we got that uh double negated r uh well let's use material implication on this then we get a disjunction right with a negation it's a double negation but still negation yeah so now that's equivalent to if not r then if not q then not p well now we can use exportation that's how we get that uh conjunction in the antecedent use exportation and we can join together not r and not q to infer not p all right so uh now we're seeing right where we start adding up these rules together and it can uh it can be in a lot it could be a lot a lot so just keep this in mind when using these rules of equivalence keep this in mind without the rules of implication it's rarely going to be a single step from the premises to the conclusion you probably have to use many steps in between right so you have to kind of chart out that path and the way you do that is you look at the conditional excuse me look at the uh conclusion find the conclusion and the premises that's probably not going to give you the whole story one part of the story then look at the premises and see which rules of application and which equivalence rules are going to be applicable to those premises uh to get to that conclusion you kind of have to work both directions from conclusion and premises and from premises to conclusion all right let's keep going okay so we've had one equivalence rule with negations but it wasn't really mixing with much else uh now let's look at an equivalence rule with conjunctions and equivalence rule with disjunctions right now suppose we have a negation of a conjunction right now you might think that you could just like you think mathematics right well it's negated so i can negate every number within the parentheses that doesn't work that way it's not the same thing uh the negation is spread out over both conjuncts right but instead of a conjunction now you have a disjunction so when you negate a conjunction you think it's false at both of these it's false that uh i am outside and i am eating lunch it's false at both of those uh well that just means that at least one of those is false right it's you know it's false that i'm eating lunch i'm outside but it's false that i'm eating lunch so when you negate a conjunction what that's equivalent to a disjunction where both disjuncts are negated and and vice versa if you have two disjuncts that are negated that's equivalent to a conjunction where the entire conjunction is negated right okay now if you have a negation of a disjunction right disjunction says at least one of these is true well if that's negated then both are false both are false so if you have a negation of a disjunct that's equivalent to a conjunction where each conjunct is negated and if you have a conjunction where each conjunct is negated that's equivalent to a disjunct that's that a disjunction that's negated Okay, so let's take a look at some problems with this. This is demorgans, demorgans. Okay, so we've got p, we've got, we've got not p and we've got not q both as two premises and the third premise is if r then p or q. Okay, so already this looks suspicious because now we're pulling r apparently out of nowhere, right, and we want to, the conclusion is not r. Well, that conclusion isn't, well, it's found, sorry, it's not out of nowhere. It's found in the antecedent of the third premise, sorry, it's found in the antecedent of the third premise. Okay, well if it's the antecedent of the third premise, chances are we're using modus tollens and use modus tollens what we, what we need is not p or q, right, we've got to negate the consequences, not p or q. Well, how do we get that? Because we don't have not p or q, it's false that p or q, I should say, it's false that p or q in, as in the premises, but we've got not p and not q separately. Okay, well, let's conjoin them. So we'll take not p and not q separately, we'll conjoin them together as a good junction and introduction, then using demorgans, that's how we're able to infer the negation of the disjunction and with the negation of the disjunction, right, or actually let's just be fancy, right. So I said modus tollens, you can use modus tollens, but since I've shown up equivalence rules, let's use the equivalence rule on that conditional and swap it around. So we got if not p or q, then implies not r. So we infer not r. There's lots of different ways to do these proofs. You could use modus tollens, but I want to show off the equivalence rules here, so that's why I do that. All right, let's take a look at another problem. All right. All right. So we've got not p as a premise, right. Then I got a conditional r implies p and q, and the conclusion is not r, right. Conclusion is not r. Okay. Well, where's r? r is the antecedent of a conditional. So we could use modus tollens, or as we just saw, can use contraposition, contraposition. So we got not p, but we don't have q anywhere. Well, where are we going? If we don't have q anywhere, where are we going to get chances are you have to pull it out of the air, right? So we use disjunction introduction with that p. We got not p or not q. And now we use demorgans to get the negation of the conjunction. And now we got the negation of the conjunction, use demorgans. So then we use contraposition again on line two. Again, I'm showing off the equivalence rules. Could use modus tollens, but I'm showing off the equivalence rules. And then using modus ponens, lines four and five, now can infer not r. Okay, last problem. We've got not p disjoint to a disjunction of negations. That's the first premise, not q or r. Second premise is s or r. If s then r, excuse me. And now we're supposed to apply not p and q disjoint to not s. Oh boy. There's a lot going on here. Well, if we've got a negation, look at that conclusion. If you've got a negation, excuse me, a disjunction of negations, might be using what, disjunction introductions, a possibility. Might be using dilemma. Might be, now we might be using demorgans, right? Demorgans allows us to deal with disjunctions as well. Well, let's just keep all these options available to us. Let's just take a look. So, not p and q as one disjunct and not s as the other. Where's that found in the premises? Well, s is the antecedent of a conditional. Well, that might be useful. And if it's the antecedent of a conditional, we're looking for the negation of the antecedent. Well, modus tolens or a contraposition might be in our future. Either one would work. And the other one, we've got a large disjunction. We've got not p and q. Just the negation of p and q is another disjunct. Well, it probably might be demorgans, right? Because you can't conjoin two negations to get an negation of a conjunction. So, it might be demorgans. All right. Well, let's try it out. What sort of disjunction would allow us to get this disjunction? So, dilemma is probably in our future here. Dilemma is probably in our future. Since s, since all the components in the conclusion already contain in the premises, we're probably not going to need disjunction introduction. So, what can we do with that? Well, let's try and get the right kind of disjunction. Association is in our future. We want p and q together on one side and when s on the other. Well, then let's take that first premise and through association get p and q as one half of a disjunct and s as the other half. Excuse me, r, excuse me, r as the other half, right? All right. So, let's go ahead and use dilemma then. Let's assume not p or not q for the left-handed part of the disjunction. And from that, we can use demorgans to infer the negation of the conjunction, not p and q. Let's use not r for the right-handed part. Assume not r for the right-handed part of the disjunction. And keep this in mind with dilemma. If you're going to assume one half of the dilemma, you got to assume the other half, right? Remember, you're trying to infer something from both halves, all right? So, we got not r for the right-handed part of the disjunction. Well, we got that handy little conditional there. Let's go ahead and flip it around using contraposition again. This is showoff contraposition. We've got if s then r. Well, let's turn that into if not r, then not s. So, using that not r in that conditional, then we can infer s, right? Then we can infer s. So, now we have our disjunction that we're looking for the conditional using dilemma. That's going to, that we have in the conclusion, using dilemma. So, we got the negation of not p and q, excuse me, not p and q disjoint to not s. And how do we get that? Well, look at that. Look at the citation. Line three is where we've got the disjunction. Line four is where we assume the left-handed disjunct. Line five is where we get our conclusion for the left-handed disjunct. At least the left-handed part of the conclusion for the disjunct. Line six is where we assume the right-handed portion of the disjunct. And then line eight is where we infer the right-handed disjunct and the conclusion. So, that we got, put that all together, citing our infant's rule with dilemma. All right. So, that's the last equivalence rule. Let me check my pocket. Yep, that's the last equivalence rule. I think that's all of it. Yeah. So, you know, with this set of problems, you'll be using combination of equivalence rules and rules of implication. I mean, sometimes, you know, for the test it might be just implication, might be just, but you should always think about the possibility of a combination. And when you're doing these problems, right, so when I give them to you, either the formula or filled out or the rules are filled out, so that's already half the work, right? You're gonna have to work for you. You just have to look at, if the formulas are filled out, you have to look and say, okay, which rules get you and which lines get you that formula and the rules are filled out, then you have to figure, okay, what can I infer using these rules and these lines, right? Okay, so that's the rules, that's the equivalence rules. Probably take a look at some homework problems.