Theorem equsal 2408

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2363. See equsalvw 1999 and equsalv 2250 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsex 2409. (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 2196	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
3		equsal.2	. . . 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		ax6e 2374	. . 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  ax-13 2363

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-nf 1778

This theorem is referenced by:  equsex  2409  equsalh  2411  dvelimf  2439  sb6x  2455  sb6rf  2459  bj-sbievv  36217