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 𝑥 𝐴
spcgf.2 𝑥 𝜓
spcgf.3 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion spcgf ( 𝐴𝑉 → ( ∀ 𝑥 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 spcgf.1 𝑥 𝐴
2 spcgf.2 𝑥 𝜓
3 spcgf.3 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
4 2 1 spcgft ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝐴𝑉 → ( ∀ 𝑥 𝜑𝜓 ) ) )
5 4 3 mpg ( 𝐴𝑉 → ( ∀ 𝑥 𝜑𝜓 ) )