Posted by: hrummel | December 1, 2008

X implies Y is true when X is false, Y is true.

The logic statement X implies Y is true when X is false and Y is true has always confused me. When spoken it seems strange that “the Germans won WW2 implies that rabbits are cute” is a true statement. It seems perfectly logical X implies Y is false when X is true and Y is false. “rabbits are cute implies that the Germans won WW2” is a false statement. However when X is false, we are not given any information about how it relates to Y. Thus mathematicians have chosen to assign truth to this statement when X is false. The result is, when looking at a Venn diagram, that we have a truth value everywhere except when A is true and B is false.

As to why mathematicians like having X implies Y to be true when X is false, we have simple representations for all other causes. We definitely wish it to be false when X is true and Y is false. This means that a true statement would imply a false statement, which should not happen. 1+1=2 should not imply 1+1=3. When looking at a Venn diagram, the statements that make sense are when X is true, and Y is false, the statement is false, when X is true and Y is true, the statement is true. Suppose we were to change the truth table such that X false, Y false, X->Y false then we would arrive at the representation “Y is true”.

Suppose we were to change the truth table such that X false, Y true, X->Y false then we would arrive at the representation “X if and only if Y”

Suppose we were to change the truth table such that anytime X is false, X->Y is false, we would arrive at the representation for “X and Y”.

Thus choosing “X implies Y” to be true anytime X is false we arrive at a new truth table that is not constructed with another easy expression, and we avoid redundancy. This is most likely why mathematicians have chosen the truth values that exist.

Responses

1. aww… you think that rabbits are cute!

pretty Venn diagrams, too.

2. I like this explanation. It is probably one of the reasons, I agree.

3. Suppose “X implies Y is false when X is false and Y is true” is a theorem of our logic L. i.e. (X & ~Y) -> ~(X -> Y). Then (X -> Y) -> ~(X & ~Y), modus ponens. Using de Morgan’s laws, the RHS is equivalent to ~X & Y. Then both ((X -> Y) -> ~X) and ((X -> Y) -> Y)). But both of these statements run contrary to our intuitive understanding of implication! The first says that, “If Y follows from X, then X is untrue”. The second says that “If Y follows from X, then Y is true”.

The basic problem, then with NOT assuming that “X implies Y is true when X is false and Y is true”, is that we are led to a contradiction. I have only sketched a proof of this assertion, but believe it can be made rigorous.

Do you see any difficulties with this line of reasoning?

4. So very true.

5. Use CKod (http://ckod.sourceforge.net/_/) for creating your “truth table”.