Theorem equsalv 2250

Description: An equivalence related to implicit substitution. Version of equsal 2410 with a disjoint variable condition, which does not require ax-13 2365. See equsalvw 1999 for a version with two disjoint variable conditions requiring fewer axioms. See also the dual form equsexv 2251. (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 2196	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
3		equsalv.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	pm5.74i 271	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜓))
5	4	albii 1813	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
6		ax6ev 1965	. . 3 ⊢ ∃𝑥 𝑥 = 𝑦
7	6	a1bi 362	. 2 ⊢ (𝜓 ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
8	2, 5, 7	3bitr4i 303	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1531  ∃wex 1773  Ⅎwnf 1777

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-12 2163

This theorem depends on definitions:  df-bi 206  df-ex 1774  df-nf 1778

This theorem is referenced by:  equsexv  2251  equsexvOLD  2252  equsalhw  2279  sbiev  2302  sb6rfv  2347  nfabdwOLD  2921  bj-equsalhv  36191  ichnfimlem  46685