 We will discuss methods of proof of an implication so far we have used truth tables to specify propositional functions including the basic functions involving a single logical connectives we have also encountered tautologies which are foundations upon which valid inference are best now we shall apply valid inference patterns to validate nine common methods for proving implications these will be required these will be referred to as methods of proof so the first method of proof is trivial proof p implying q it is possible to establish that q is true regardless of the truth value of p then p implying q is true this happens because if we see the truth table of p implies q we see that when p is f and q is f that is p and q both are false then p implies q is true when p is false and q is true then p implies q is true when p is true and q is false then p implying q is true and when p is true and q is true then p implies q is true this means that if it so happens that we know that q is always true then p implies q is true regardless of what happens to p this is of course trivial and hence it is called the trivial proof the second method of proof is vacuous proof if it is possible to establish that q is true I will reward this so here if p is shown to be false regardless of the truth value of q then also p implying q is true if p is shown to be false regardless of the truth value of q then p implies q is true we see this if we again look at the truth table of p implies q we see that this portion of the truth table says that p implies q is true if p is fixed to f that is p is false whatever be the values of q next we move on to direct proof of p implies q the construction of a direct of p implies q begins by assuming that p is true then from the available information from the reference frame the conclusion q is shown to be true by valid inference so direct proof is indeed direct we have to prove p implies q so we will assume that the proposition p is true and then use the valid inferences and the information available in the reference frame and to step by step deduce that q is true if q is true whenever p is true then the as proposition p implies q is true the fourth method of proof is indirect proof of p implies q this is basically the direct proof of the contrapositive we recall that if we have a proposition p implies q then contrapositive of that proposition is not of q implies not of p and p implies q and not of q implies not of p are equivalent statements therefore if we can prove that negation of q implies negation of p then we have proved p implies q we need to work out this proof in two steps I am coming to that indirect proof proceeds as follows a assume q is false b on the basis of that assumption the available information from the frame of reference p is false 5 of p implies q by contradiction this method exploits the fact p implies q is true if p and negation of q is false now we have seen that p implies q is equivalent to the statement negation of p or q if we take negation of p or q and negation of that then this is equivalent to negation of negation of p and negation of q this is by De Morgan's law and which in turn is equivalent to p and negation of q therefore if we can prove that p and negation of q is false then the negation of that which comes from here that is p negation of p or q is true conversely the proof by contradiction is a very powerful tool and to implement such a proof we have to go step wise now I write the steps of the proof by contradiction a assume that p and q negation of q is false assume that this is true not false in the contrary start by assuming that p and q complement that is not negation of q is true step b discover on the basis of that assumption some conclusion that is patently false see then the contradiction in the step b leads us to conclude that the assumption a is false then the contradiction in step b leads us to conclude that the assumption in step a is false in step a we assumed that p and negation of q is true so that means that p and negation of q is false this will imply that p negation of q negation of that that is negation of p or q which is in turn equivalent to p implies q is true this is what is known as the proof by contradiction the sixth method is proof of p implies q by cases now here we are in a situation where p can be split up into or of several propositions p1 p2 up to pn and we are encountering a proposition like this which is p1 or p2 or and so on up to or pn implies q where the left hand side is essentially p now what we note that we can establish this fact by proving smaller implications such as p1 implies q p2 implies q and pn implies q if all of them are true then this is also true the seventh method is proof by elimination of cases if we are confronted alternatives to be true as to be true we are able to verify p is false then we must conclude that q is true this method of proof is essentially disjunctive syllogism from another perspective if pq are two propositions then disjunctive syllogism is as follows here we see that in the two propositions if p or q is true and if we know that p is not true then that will imply q is true we can extend this for finite number of cases as follows if p1 up to pn are n propositions then p1 or p2 or q p1 negation p2 negation and and so on pn negation this is q is a tautology next the two propositions p implies q implying r and p and q implies r are equivalent therefore p implies q implies r can be as follows combine the two antecedents p and q b then prove r on the basis of the assumption now we have come to the last method of proof that we discuss in this lecture proof of equivalence now suppose we have a biconditional p if and only if q and we would like to prove it proof of equivalence says that it is enough to prove separately the direct implication p implies q and its converse q implies p and this is proof of implication now that we have discussed some methods of proofs of implications we must also remember that these are not all the possible methods there are other methods as well but these are the methods which are very commonly used and it is a good idea to try to formulate the proofs based on these methods we will now look at examples of proofs constructed by using proof by contradiction and the proof by contrapositive so first let us check proof by contradiction suppose that 10 integers 1 to up to 10 are randomly positioned a circle or a circular wheel show that the sum of set of three consecutively positioned numbers is at least 17 so I read again now suppose I have got 10 integers 1 to up to 10 and I have a circular wheel with 10 positions and at each position I am putting some number may be 1 2 4 and so on like that what I am claiming is that there is a there is three consecutive positions such that if I add up the numbers placed in those three consecutive positions then it will be at least 17 no matter in which way I arrange the numbers between 1 to 10 around the circular wheel so we start the proof by defining Xi which is the integer positioned at the position I of the wheel now we have to prove that either x1 plus x2 plus x3 greater than or equal to 17 or x2 plus x3 plus x4 greater than or equal to 17 and so on lastly or x10 plus x1 plus x2 greater than or equal to 17 so we want to prove this we want to prove that one of at least one of these one of these conditions must hold now we prove by contradiction so we say that let us suppose that it does not hold so if it does not hold then what will hold we will have if the above is not true then x1 plus x2 plus x3 less than 17 please check that it is strict inequality then x2 plus x3 plus x4 this is also less than 17 and so on up to x10 plus x1 plus x2 less than 17 and this is and all of this must hold because if one of them is not true then that sum is greater than or equal to 17 and that means my proposition is correct so suppose this is true now we add up all these inequalities to obtain three times so we are assuming that this is true so that means a three times x1 plus and so on up to x10 is less than 10 less than or equal to 10 into 16 which is equal to 160 now we remember that x1 up to x10 are distinct integers between 1 to 10 so we can add them up to obtain 10 into 11 divided by 2 into 3 which gives me 165 and if our assumption is true then 165 is less than or equal to 160 which is of course false so this is a contradiction but all the logical inferences that we have used are valid therefore through a valid logical inference we are finding that truth is imply something false that means that the premise cannot be true and therefore from this we can conclude that our original assumption is true that is there are consecutive three numbers always whose sum is greater than or equal to 17 this is an example of a proof by contradiction as a last topic of today's lecture we will discuss an example of a proof by contrapositive the statement that we would like to prove is if the product of two integers a and b is even then either a is even or b is even now we denote the antecedent by the proposition p that is a b is even and q denotes the consequent a is even or b is even we want to prove that p implies q we start with not of q not of q is a is odd and b is odd this is by using de Morgan's laws so we know that a is we are assuming that a is odd and b is odd therefore a can be written as 2m plus 1 and b can be written as 2n plus 1 if it take the product a b then this is 2m plus 1 into 2n plus 1 which gives me 2 times 2mn plus m plus n plus 1 now we see that a plus b is odd sorry a into b is odd that is the product of a b is odd therefore negation of p is true thus we have proved that if negation of q is true that is whenever negation of q is true negation of p is true therefore we have proved the implication negation of q implies negation of p which is equivalent to p implies q this is proof by using contract positive and this is the end of today's lecture thank you.