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Metamath Proof Explorer

Theorem funALTVeqd

Description: Equality deduction for the function predicate. (Contributed by NM, 23-Feb-2013)

		Ref	Expression
	Hypothesis	funALTVeqd.1	$⊢ φ \to A = B$
	Assertion	funALTVeqd	$⊢ φ \to (FunALTV A \leftrightarrow FunALTV B)$

Step	Hyp	Ref	Expression
1		funALTVeqd.1	$⊢ φ \to A = B$
2		funALTVeq	$⊢ A = B \to (FunALTV A \leftrightarrow FunALTV B)$
3	1 2	syl	$⊢ φ \to (FunALTV A \leftrightarrow FunALTV B)$