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Metamath Proof Explorer

Theorem axc5c7toc5

Description: Rederivation of ax-c5 from axc5c7 . Only propositional calculus is used for the rederivation. (Contributed by Scott Fenton, 12-Sep-2005) (Proof modification is discouraged.) (New usage is discouraged.)

		Ref	Expression
	Assertion	axc5c7toc5	$⊢ \forall x φ \to φ$

Proof

Step	Hyp	Ref	Expression
1		ax-1	$⊢ \forall x φ \to (\forall x \neg \forall x φ \to \forall x φ)$
2		axc5c7	$⊢ (\forall x \neg \forall x φ \to \forall x φ) \to φ$
3	1 2	syl	$⊢ \forall x φ \to φ$