Metamath Proof Explorer


Theorem spv

Description: Specialization, using implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker spvv if possible. (Contributed by NM, 30-Aug-1993) (New usage is discouraged.)

Ref Expression
Hypothesis spv.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion spv ( ∀ 𝑥 𝜑 → 𝜓 )

Proof

Step Hyp Ref Expression
1 spv.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2 1 biimpd ⊢ ( 𝑥 = 𝑦 → ( 𝜑 → 𝜓 ) )
3 2 spimv ⊢ ( ∀ 𝑥 𝜑 → 𝜓 )