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Kind of a weird question. If propositions $P$ and $Q$ are equivalent and if we can show that, for some other proposition $R$, $P \implies R$ without using $Q$, then is there a proof of $Q \implies R$ that doesn't use $P$?

My thought is that if every proof of $Q \implies R$ ends up using $P$, then maybe $P$ and $Q$ are not equivalent after all.

Context: Part III here: Finding an almost complex structure (aka anti-involution) given an involution

BCLC
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    "It is possible to do without" is not the same as "It is necessary to do without" – Graham Kemp Mar 07 '20 at 04:56
  • @GrahamKemp thanks. edited post – BCLC Mar 07 '20 at 05:33
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    Let $Q = \text{True }\wedge P$ and $R = \text{False } \vee P$. You: "Oh no! You can't prove $Q \implies R$ without using $P$! Obviously $Q$ is not really equivalent to $P$!". Me: "How closely have you been monitoring your medications?" – Paul Sinclair Mar 07 '20 at 15:22
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    @PaulSinclair Lol ok thanks! – BCLC Apr 30 '20 at 14:19
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    You are welcome. I'm glad you took it in the spirit intended. – Paul Sinclair Apr 30 '20 at 15:59
  • @PaulSinclair Related? https://math.stackexchange.com/questions/3626912/proving-a-thorem-using-its-converse – BCLC Nov 23 '20 at 12:56
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    I don't see how. Your question has nothing to do with the converse relationships. And in the other thread José is not expressing any doubts about the validity of the proof (the Pythagorean theorem has been proved before, so there is no logical inconsistency in using this statement to prove its converse). He just considered it an interesting approach and is looking for other examples. I am certain I have seen other examples myself, but I cannot recall exactly what they were. – Paul Sinclair Nov 23 '20 at 17:40
  • @PaulSinclair thanks! – BCLC Dec 03 '20 at 23:35

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