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Metamath Proof Explorer

Theorem axc16

Description: Proof of older axiom ax-c16 . (Contributed by NM, 8-Nov-2006) (Revised by NM, 22-Sep-2017)

		Ref	Expression
	Assertion	axc16	$⊢ \forall x x = y \to (φ \to \forall x φ)$

Step	Hyp	Ref	Expression
1		axc16g	$⊢ \forall x x = y \to (φ \to \forall x φ)$