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Metamath Proof Explorer

Theorem equsalhw

Description: Version of equsalh with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 29-Nov-2015) (Proof shortened by Wolf Lammen, 8-Jul-2022)

	Ref	Expression
Hypotheses	equsalhw.1	⊢ ( 𝜓 → ∀ 𝑥 𝜓 )
	equsalhw.2	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion	equsalhw	⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		equsalhw.1	⊢ ( 𝜓 → ∀ 𝑥 𝜓 )
2		equsalhw.2	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
3	1	nf5i	⊢ Ⅎ 𝑥 𝜓
4	3 2	equsalv	⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ 𝜓 )