Theorem equsalhw 2225

Description: Version of equsalh 2355 with a disjoint variable condition, which does not require ax-13 2301. (Contributed by NM, 29-Nov-2015.) (Proof shortened by Wolf Lammen, 8-Jul-2022.)

Hypotheses

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

Assertion

Ref	Expression
equsalhw	⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem equsalhw

Step	Hyp	Ref	Expression
1		equsalhw.1	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
2	1	nf5i 2084	. 2 ⊢ Ⅎ𝑥𝜓
3		equsalhw.2	. 2 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	2, 3	equsalv 2196	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 198  ∀wal 1505

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-10 2079  ax-12 2106

This theorem depends on definitions:  df-bi 199  df-ex 1743  df-nf 1747

This theorem is referenced by:  dvelimhw  2279