Metamath Proof Explorer


Theorem spcev

Description: Existential specialization, using implicit substitution. (Contributed by NM, 31-Dec-1993) (Proof shortened by Eric Schmidt, 22-Dec-2006)

Ref Expression
Hypotheses spcv.1 𝐴 ∈ V
spcv.2 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion spcev ( 𝜓 → ∃ 𝑥 𝜑 )

Proof

Step Hyp Ref Expression
1 spcv.1 𝐴 ∈ V
2 spcv.2 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
3 2 spcegv ( 𝐴 ∈ V → ( 𝜓 → ∃ 𝑥 𝜑 ) )
4 1 3 ax-mp ( 𝜓 → ∃ 𝑥 𝜑 )