Theorem equsal 2446

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2401. See equsalvw 2036 and equsalv 2301 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsex 2447. (Contributed by NM, 2-Jun-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof shortened by Wolf Lammen, 5-Feb-2018.) (New usage is discouraged.)

Hypotheses

Ref	Expression
equsal.1	⊢ Ⅎ𝑥𝜓
equsal.2	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

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

Proof of Theorem equsal

Step	Hyp	Ref	Expression
1		equsal.1	. . 3 ⊢ Ⅎ𝑥𝜓
2	1	19.23 2247	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
3		equsal.2	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	pm5.74i 274	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜓))
5	4	albii 1851	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
6		ax6e 2412	. . 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 1567  ∃wex 1811  Ⅎwnf 1815

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-12 2212  ax-13 2401

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816

This theorem is used by:  equsex  2447  equsalh  2449  dvelimf  2477  sb6x  2493  sb6rf  2497  bj-sbievv  37852