Theorem rexprg 4701

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 2129, ax-12 2166. (Revised by Gino Giotto, 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 317	. . 3 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
3		ralprg.2	. . . 4 ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒))
4	3	notbid 317	. . 3 ⊢ (𝑥 = 𝐵 → (¬ 𝜑 ↔ ¬ 𝜒))
5	2, 4	ralprg 4699	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)))
6		ralnex 3062	. . . 4 ⊢ (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑)
7		pm4.56 986	. . . 4 ⊢ ((¬ 𝜓 ∧ ¬ 𝜒) ↔ ¬ (𝜓 ∨ 𝜒))
8	6, 7	bibi12i 338	. . 3 ⊢ ((∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
9		notbi 318	. . 3 ⊢ ((∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
10	8, 9	sylbb2 237	. 2 ⊢ ((∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)))
11	5, 10	syl 17	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 394   ∨ wo 845   = wceq 1533   ∈ wcel 2098  ∀wral 3051  ∃wrex 3060  {cpr 4631

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-ral 3052  df-rex 3061  df-v 3465  df-un 3950  df-sn 4630  df-pr 4632

This theorem is referenced by:  rextpg  4704  rexpr  4706  reurexprg  4709  fr2nr  5655  sgrp2nmndlem5  18886  nb3grprlem2  29251  nfrgr2v  30139  3vfriswmgrlem  30144  brfvrcld  43203  rnmptpr  44631  ldepspr  47669  zlmodzxzldeplem4  47699