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Metamath Proof Explorer

Theorem fveq1d

Description: Equality deduction for function value. (Contributed by NM, 2-Sep-2003)

		Ref	Expression
	Hypothesis	fveq1d.1	$⊢ φ \to F = G$
	Assertion	fveq1d	$⊢ φ \to F (A) = G (A)$

Step	Hyp	Ref	Expression
1		fveq1d.1	$⊢ φ \to F = G$
2		fveq1	$⊢ F = G \to F (A) = G (A)$
3	1 2	syl	$⊢ φ \to F (A) = G (A)$