Metamath Proof Explorer


Theorem spcgv

Description: Rule of specialization, using implicit substitution. Compare Theorem 7.3 of Quine p. 44. (Contributed by NM, 22-Jun-1994) Avoid ax-10 , ax-11 . (Revised by Wolf Lammen, 25-Aug-2023)

Ref Expression
Hypothesis spcgv.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion spcgv ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 𝜑 → 𝜓 ) )

Proof

Step Hyp Ref Expression
1 spcgv.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 elex ⊢ ( 𝐴 ∈ 𝑉 → 𝐴 ∈ V )
3 elex ⊢ ( 𝐴 ∈ V → 𝐴 ∈ V )
4 1 adantl ⊢ ( ( 𝐴 ∈ V ∧ 𝑥 = 𝐴 ) → ( 𝜑 ↔ 𝜓 ) )
5 3 4 spcdv ⊢ ( 𝐴 ∈ V → ( ∀ 𝑥 𝜑 → 𝜓 ) )
6 2 5 syl ⊢ ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 𝜑 → 𝜓 ) )