Theorem rexprg 4649

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 2146, ax-12 2182. (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 318	. . 3 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
3		ralprg.2	. . . 4 ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒))
4	3	notbid 318	. . 3 ⊢ (𝑥 = 𝐵 → (¬ 𝜑 ↔ ¬ 𝜒))
5	2, 4	ralprg 4648	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)))
6		ralnex 3059	. . . 4 ⊢ (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑)
7		pm4.56 990	. . . 4 ⊢ ((¬ 𝜓 ∧ ¬ 𝜒) ↔ ¬ (𝜓 ∨ 𝜒))
8	6, 7	bibi12i 339	. . 3 ⊢ ((∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
9		notbi 319	. . 3 ⊢ ((∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
10	8, 9	sylbb2 238	. 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 206   ∧ wa 395   ∨ wo 847   = wceq 1541   ∈ wcel 2113  ∀wral 3048  ∃wrex 3057  {cpr 4577

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-rex 3058  df-v 3439  df-un 3903  df-sn 4576  df-pr 4578

This theorem is referenced by:  rextpg  4651  rexpr  4653  reurexprg  4656  fr2nr  5596  sgrp2nmndlem5  18839  nb3grprlem2  29361  nfrgr2v  30254  3vfriswmgrlem  30259  brfvrcld  43808  rnmptpr  45298  ldepspr  48598  zlmodzxzldeplem4  48628