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Hello this might sound like a play on word although I am kinda of confused. Can it be possible that two propositions are both tautologies but not logically equivalent?

Here is an example:

$(\neg p \wedge (p \vee q)) \rightarrow p$ and $(p \wedge(p\rightarrow q)) \rightarrow q$ now these two are both tautologies, and if I try to show that they are with identities they are both true at the end where:

$(p \vee T)$ and $(q \vee T)$ (Thus both true)

but when you try to show logical equivalence with identities, then you can't.

Is this right? and if yes why? How is this called?

Thank you

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Any two tautologies are logically equivalent.

And note: $p \lor \top = q \lor \top = \top$

So they really are equivalent!

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  • $\begingroup$ But then all tautologies are logically equivalent between each other? $\endgroup$ – The Law Feb 4 '17 at 23:55
  • $\begingroup$ Correct!! You can rewrite any tautology as $\top$. $\endgroup$ – Bram28 Feb 4 '17 at 23:55
  • $\begingroup$ Ah ok, now it makes sense, thanks! $\endgroup$ – The Law Feb 4 '17 at 23:58
  • $\begingroup$ @Bram28 If all tautologies are logically equivalent; then why do we have different axioms that are tautologies? See math.stackexchange.com/questions/2119044/… $\endgroup$ – user400188 Feb 5 '17 at 0:14
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    $\begingroup$ @user400188 Excellent question!! The answer is that for these kinds of axiom systems we don't have the rules of equivalence available to us to actually derive one from the others ... In fact, those systems typically have just one rule of inference: Modus Ponens. But yes, if we did have the equivalence rules available to use you would be absolutely right: we would need just one axiom instead of a bunch! $\endgroup$ – Bram28 Feb 5 '17 at 0:34

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