What's the proper way to prove P$\rightarrow$Q is not tautologically equivalent to Q$\rightarrow$P

Question

I want to rigorously show that [$(p \rightarrow q)\equiv(q \rightarrow q)$] is not true.

$Let A =(p \rightarrow q), B =(q \rightarrow q)$

I tried to prove this by supposing $A$ and showing that $B$ does not follow.

[$(p \rightarrow q) \rightarrow (q \rightarrow p)$]

I expanded that but it didn't $= 0$

I was told to look at Reductio Ad Absurdum, but I don't see how that applies in this case.

This is a real-world example. This is not homework.

Answer 1

The most straightforward way is to construct the truth tables and see that they are not equal. Or you could show values of $P$ and $Q$ that make them different (which would amount to mention the specific line of the truth tables where they differ).

If $P$ is true and $Q$ is false, then $P\to Q$ is false, while $Q\to P$ is true.

Edit: Changed $Q\to Q$ to $Q\to P$, to go with the title.

Answer 2

Assume $B$. Show that $A$ could be false.

By a similar method you can show that if B holds true, then (p $\rightarrow$ r) could be false.