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$\begingroup$ Lemma mod_mult_or : forall a b c, (a | b) \/ (a | c) -> (a | b * c). Proof. intros a b c H. destruct H as [Hb | Hc]. - apply Z.divide_mul_r. exact Hb. - apply Z.divide_mul_l. exact Hc. Qed. I tried this way but it seems to have not worked. :( $\endgroup$
– itsFrank
Commented Apr 25, 2023 at 16:30
$\begingroup$ I edited the first sub-goal of the proof. What was the issue with it? $\endgroup$
– Pierre Castéran
Commented Apr 25, 2023 at 20:10
$\begingroup$ I figured it out. $\endgroup$
– itsFrank
Commented Apr 25, 2023 at 21:15
$\begingroup$ Just used unfold rewrite exists -> tactic, thanks! $\endgroup$
– itsFrank
Commented Apr 25, 2023 at 21:17

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