Theorem equsex 2450

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2404. See equsexvw 2040 and equsexv 2306 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal 2449. See equsexALT 2451 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 482	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
4	1, 3	exlimi 2257	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
5	1, 2	equsal 2449	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)
6		equs4 2448	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7	5, 6	sylbir 238	. 2 ⊢ (𝜓 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8	4, 7	impbii 212	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812  Ⅎwnf 1816

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-12 2216  ax-13 2404

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817

This theorem is used by:  equsexh  2453  sb5rf  2499