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Metamath Proof Explorer

Theorem spcgf

Description: Rule of specialization, using implicit substitution. Compare Theorem 7.3 of Quine p. 44. (Contributed by NM, 2-Feb-1997) (Revised by Andrew Salmon, 12-Aug-2011)

	Ref	Expression
Hypotheses	spcgf.1	$⊢ \underline{Ⅎ} x A$
	spcgf.2	$⊢ Ⅎ x ψ$
	spcgf.3	$⊢ x = A \to (φ \leftrightarrow ψ)$
Assertion	spcgf	$⊢ A \in V \to (\forall x φ \to ψ)$

Proof

Step	Hyp	Ref	Expression
1		spcgf.1	$⊢ \underline{Ⅎ} x A$
2		spcgf.2	$⊢ Ⅎ x ψ$
3		spcgf.3	$⊢ x = A \to (φ \leftrightarrow ψ)$
4	2 1	spcgft	$⊢ \forall x (x = A \to (φ \leftrightarrow ψ)) \to (A \in V \to (\forall x φ \to ψ))$
5	4 3	mpg	$⊢ A \in V \to (\forall x φ \to ψ)$