Theorem equsalv 2303

Description: An equivalence related to implicit substitution. Version of equsal 2448 with a disjoint variable condition, which does not require ax-13 2403. See equsalvw 2037 for a version with two disjoint variable conditions requiring fewer axioms. See also the dual form equsexv 2304. (Contributed by NM, 2-Jun-1993.) (Revised by BJ, 31-May-2019.)

Hypotheses

Ref	Expression
equsalv.nf	⊢ Ⅎ𝑥𝜓
equsalv.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

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

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem equsalv

Step	Hyp	Ref	Expression
1		equsalv.nf	. . 3 ⊢ Ⅎ𝑥𝜓
2	1	19.23 2249	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
3		equsalv.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	pm5.74i 274	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜓))
5	4	albii 1852	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
6		ax6ev 2002	. . 3 ⊢ ∃𝑥 𝑥 = 𝑦
7	6	a1bi 365	. 2 ⊢ (𝜓 ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
8	2, 5, 7	3bitr4i 306	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  ∃wex 1812  Ⅎwnf 1816

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-12 2215

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

This theorem is used by:  equsexv  2304  equsalhw  2326  sb6rfv  2388  bj-equsalhv  37534  ichnfimlem  48348