Theorem rexprg 4664

Description: Convert a restricted existential quantification over a pair to a disjunction. (Contributed by NM, 17-Sep-2011.) (Revised by Mario Carneiro, 23-Apr-2015.) Avoid ax-10 2176, ax-12 2213. (Revised by GG, 30-Sep-2024.)

Hypotheses

Ref	Expression
ralprg.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
ralprg.2	⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒))

Assertion

Ref	Expression
rexprg	⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥   𝜒,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)   𝑊(𝑥)

Proof of Theorem rexprg

Step	Hyp	Ref	Expression
1		ralprg.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
2	1	notbid 321	. . 3 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
3		ralprg.2	. . . 4 ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒))
4	3	notbid 321	. . 3 ⊢ (𝑥 = 𝐵 → (¬ 𝜑 ↔ ¬ 𝜒))
5	2, 4	ralprg 4663	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)))
6		ralnex 3091	. . . 4 ⊢ (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑)
7		pm4.56 1004	. . . 4 ⊢ ((¬ 𝜓 ∧ ¬ 𝜒) ↔ ¬ (𝜓 ∨ 𝜒))
8	6, 7	bibi12i 342	. . 3 ⊢ ((∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
9		notbi 322	. . 3 ⊢ ((∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
10	8, 9	sylbb2 241	. 2 ⊢ ((∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)))
11	5, 10	syl 18	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1570   ∈ wcel 2143  ∀wral 3079  ∃wrex 3089  {cpr 4592

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-v 3457  df-un 3911  df-sn 4591  df-pr 4593

This theorem is referenced by:  rextpg  4666  rexpr  4668  reurexprg  4671  fr2nr  5640  sgrp2nmndlem5  18992  nb3grprlem2  29712  nfrgr2v  30604  3vfriswmgrlem  30609  brfvrcld  44400  rnmptpr  45878  ldepspr  49236  zlmodzxzldeplem4  49266