Show that (p→q)→(r→s) and (p→r)→(q→s) are not logically equivalent.

Question

This is a problem in my math book, however, the answer is in the back of the book as it is an odd. What I don't understand, is the fact that if I plugin r = T and p,q,s = F I end up with...

(F→F)→(T→F)       (F→T)→(F→F)
    T→F               T→T
     F                 T

but when I plugin p = T and q, r, s = F I get the following...

    (T→F)→(F→F)       (T→F)→(F→F)
        F→T               F→T
         T                 T

Why is it logically equivalent in the second case but not in the first?

Answer 1

You don't need to be concerned with the "why" of this question anymore than you need to be concerned with why some birds can fly and some birds can't. For the two statements to be equivalent, the two statements must be equivalent for ALL truth values you assign to your parameters of $p,q,r,s$ (making sure you assign the same ones to both statements.) Since you have found a case where the two statements are not equivalent, you can deduce that the statements are not equivalent. The fact that you also found a case where the two are equivalent doesn't mean anything if you know there exists a case where the two are not equal.

Answer 2

Consider the expressions $$a + (2\times b)$$ and $$(a\times b) + 3$$ and suppose you were asked to show that these are not equivalent arithmetically. You can put in $2$ for $a$ and $3$ for $b$ and find that the first expression has the value $2+(2\times 3) = 8$ while the second has $(2\times 3) + 3 = 9$. Then you can conclude that they are not equivalent, because equivalent expressions always have the same value no matter what you put in for the variables.

But suppose you also noticed that if you put in $1$ for $a$ and $2$ for $b$ then both expressions have the value $5$. "Why are the equivalent in one case and not in the other", you ask?

They are still not equivalent; they just happen to have the same value when you put in $1$ for $a$ and $2$ for $b$. Equivalent expressions always have the same value, and these sometimes don't, so these are not equivalent.

Or consider this analogy. Suppose you know that $A$ and $B$ are people, perhaps the same and perhaps different; you don't know who they are. But you know that $A$ had five dollars in his wallet on January 3rd while $B$ had only three dollars that day. Then you can be sure they are different people, because if they were the same person they would have had the same amount of money. But $A$ and $B$ can still be different people even if you also know that they happened to be carrying the same amount of money on a different day.