Metamath Proof Explorer


Theorem equsalvw

Description: Version of equsalv with a disjoint variable condition, and of equsal with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw . (Contributed by BJ, 31-May-2019)

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

Proof

Step Hyp Ref Expression
1 equsalvw.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2 1 pm5.74i ⊢ ( ( 𝑥 = 𝑦 → 𝜑 ) ↔ ( 𝑥 = 𝑦 → 𝜓 ) )
3 2 albii ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) )
4 equsv ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) ↔ 𝜓 )
5 3 4 bitri ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ 𝜓 )