Theorem equsex 2426

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2380. See equsexvw 2004 and equsexv 2269 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal 2425. See equsexALT 2427 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 2218	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
5	1, 2	equsal 2425	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)
6		equs4 2424	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7	5, 6	sylbir 235	. 2 ⊢ (𝜓 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8	4, 7	impbii 209	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1535  ∃wex 1777  Ⅎwnf 1781

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-12 2178  ax-13 2380

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-nf 1782

This theorem is referenced by:  equsexh  2429  sb5rf  2475