Theorem equsex 2413

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2367. See equsexvw 2001 and equsexv 2255 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal 2412. See equsexALT 2414 for an alternate proof. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof shortened by Wolf Lammen, 6-Feb-2018.) (New usage is discouraged.)

Hypotheses

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

Assertion

Ref	Expression
equsex	⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Proof of Theorem equsex

Step	Hyp	Ref	Expression
1		equsal.1	. . 3 ⊢ Ⅎ𝑥𝜓
2		equsal.2	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	biimpa 476	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
4	1, 3	exlimi 2206	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
5	1, 2	equsal 2412	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)
6		equs4 2411	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7	5, 6	sylbir 234	. 2 ⊢ (𝜓 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8	4, 7	impbii 208	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395  ∀wal 1532  ∃wex 1774  Ⅎwnf 1778

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-12 2167  ax-13 2367

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1775  df-nf 1779

This theorem is referenced by:  equsexh  2416  sb5rf  2462