Hi, suppose I have the following theorem:
th = |- !x. P x = Q x How can I convert it into this one? |- !x. P x ==> Q x As the first step, I tried "ONCE_REWRITE_RULE [EQ_IMP_THM] th", and got this: |- !x. (P x ==> Q x) /\ (Q x ==> P x) But then I don’t know how to benefit from AND1_THM (|- ∀t1 t2. t1 ∧ t2 ⇒ t1), or there’s a better method to do the whole thing? Regards, Chun Tian
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