Question about logic implications

Question

If $X\implies Y$ and $X\implies Z$, does that mean that $Y\implies Z$?

I think it does, but can anyone show this as a proof?

Thanks

As an example (x = 4) -> (x is even), and (x = 4) -> (x > 3), but (x is even) -> (x > 3) is not true. — Davis Yoshida, Mar 05 '14 at 05:51

score 1 · Accepted Answer · answered Mar 05 '14 at 06:02

This is not true. Assume for purposes of contradiction that $X\implies Y$ and $X\implies Z$ means $Y \implies Z$.

$n= 3 \implies$ $n$ is prime.

$n = 3 \implies n$ is odd.

By our assumption, $n$ is prime $\implies n$ is odd. However, $2$ is an even prime.

This is a contradiction, so our initial assumption must be false.

score 0 · Answer 2 · edited Mar 07 '14 at 13:54

0

No.

What is true is $$[(X \rightarrow Y) \land (X \rightarrow Z)]\implies (X \rightarrow (Y \land Z))$$

In the case that you know $$(X\rightarrow Y) \land (Y\rightarrow Z),$$ then you can infer $$X \rightarrow Z.$$

edited Mar 07 '14 at 13:54

Amzoti

answered Mar 06 '14 at 14:32

amWhy

2 Answers2