Theorem equsexv 2303

Description: An equivalence related to implicit substitution. Version of equsex 2449 with a disjoint variable condition, which does not require ax-13 2403. See equsexvw 2034 for a version with two disjoint variable conditions requiring fewer axioms. See also the dual form equsalv 2302. (Contributed by NM, 5-Aug-1993.) (Revised by BJ, 31-May-2019.) Avoid ax-10 2175. (Revised by GG, 18-Nov-2024.)

Hypotheses

Ref	Expression
equsalv.nf	⊢ Ⅎ𝑥𝜓
equsalv.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

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

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑥, 𝑦)

Proof of Theorem equsexv

Step	Hyp	Ref	Expression
1		equsalv.nf	. . 3 ⊢ Ⅎ𝑥𝜓
2		equsalv.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	biimpa 481	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
4	1, 3	exlimi 2252	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → 𝜓)
5	1, 2	equsalv 2302	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)
6		equs4v 2029	. . 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 400  ∀wal 1567  ∃wex 1808  Ⅎwnf 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-nf 1813

This theorem is used by:  equsexhv  2326  cleljustALT2  2396  sb10f  2558  dprd2d2  20122  poimirlem25  38324