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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 )