Theorem equsalhw 2329

Description: Version of equsalh 2455 with a disjoint variable condition, which does not require ax-13 2407. (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 2184	. 2 ⊢ Ⅎ𝑥𝜓
3		equsalhw.2	. 2 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	2, 3	equsalv 2306	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216

This proof depends on definitions:  df-bi 210  df-ex 1813  df-nf 1817

This theorem is used by:  dvelimhw  2380