 Hello and welcome to the session. In this session we discussed the following question that says draw the closure tables for the following networks represented by x plus y the whole plus x plus z the whole into x into y complement and x into y complement. What conclusion do you draw from both the polynomials? Also construct the simplified network. Let us go start with the solution. First of all let us draw the basic closure tables. These are the three basic closure tables in which we have the switches x and y. In the first table we have the switches x and y and x plus y. Now when the switches x and y are both closed which is represented by 1 then x plus y would also be 1 and in this case the current will flow. The number one represents that the current flows. So here we can say when both the switches x and y are closed then current will flow in the network x plus y. Now when the switch x is closed and y is open which is represented by 0 then x plus y would be 1 that is in this case also the current would flow. When the switch x is open and the switch y is closed then in that case x plus y would be 1 that is in this case also the current would flow. Now when x and y are both open then in that case no current will flow and so it would be represented by 0. This is the case when the switches x and y are parallel to each other. Now we consider the case when the switches x and y are in series which is represented by x into y. Now when x is 1 and y is 1 then x into y would be 1 which means that if both the switches x and y are closed then the current would flow. If the switch x is closed and y is open then no current would flow. Now when x is 0 y is 1 then x into y would be 0 that is in this case also no current would flow. When x is 0 y is 0 then x into y would be 0 that is in this case also no current would flow. So from this table we can see that the current flows only when both the switches are closed. Next we have the table in which we have x and x complement. Now when x is 1 then x complement would be 0, when x is 0 then x complement would be 1. So if this switch x is closed then x complement would be open. If this switch x is open then x complement would be closed. So these are the basic closure tables. Now the given functions are x plus y the whole plus x plus z the whole. This whole into x into y complement and the other function that we have is x into y complement. Now we will make the closure tables for both these functions. So here we have taken three switches x, y and z. Now when the three switches x, y and z are open that is we have x as 0, y as 0 and z as 0 then in that case x plus y would be 0 according to disclosure table and in that case x plus z would also be 0. Then x plus y the whole plus x plus z the whole would be this 0 plus the 0 which would be again a 0. Then y complement would be 1 as y is 0. So according to this table y complement would be 1. Then x into y complement that is the 0 into 1 would be equal to 0. So here we have 0. Now x plus y the whole plus x plus z the whole into x into y complement that is the 0 into the 0 becomes again a 0. Now next we have x 0, y 0, z 1. Then in that case x plus y would be 0, x plus z would be 1. Then x plus y the whole plus x plus z the whole would be 1. Y complement would be 1 as y is 0. Then x into y complement would be 0. x plus y the whole plus x plus z the whole into x into y complement would be 0. Next we consider x as 0, y as 1, z as 0. So x plus y would be 1, x plus z would be 0, x plus y the whole plus x plus z the whole would be 1. Y complement would be 0, x into y complement would be 0. Then x plus y the whole plus x plus z the whole into x into y complement would be 0. Next we take x as 0, y as 1, z as 1 then x plus y would be 1, x plus z would be 1, x plus y the whole plus x plus z the whole would be 1, y complement would be 0, x into y complement would be 0, x plus y the whole plus x plus z the whole into x into y complement would be 0. In the next case we take x as 1, y as 0, z as 0 then x plus y would be 1, x plus z would be 1, x plus y the whole plus x plus z the whole would be 1, y complement would be 1, x into y complement would be 1, x plus y the whole plus x plus z the whole this whole into x into y complement would be 1. Next we take x as 1, y as 0, z as 1 then x plus y would be equal to 1, x plus z would be equal to 1, x plus y the whole plus x plus z the whole would be 1, y complement would be 1, x into y complement would be 1 then x plus y the whole plus x plus z the whole into x into y complement would be 1. In the same way we have done when x is 1, y is 1 and z is 0 and when x is 1, y is 1 and z is 1. Now from this table we observe that this column and this column are equal so therefore we say that the two functions x plus y the whole plus x plus z the whole this whole into x into y complement and x into y complement are equal the networks may represent equivalent. The simplified network would be represented by this switching circuit in which we have x and y complement switches in series since these two functions are equivalent so x into y complement is represented by this which is same as this function also. So this completes the session hope you understood the solution of this question.