Theorem rexprg 4655

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 2174, ax-12 2211. (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 320	. . 3 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
3		ralprg.2	. . . 4 ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒))
4	3	notbid 320	. . 3 ⊢ (𝑥 = 𝐵 → (¬ 𝜑 ↔ ¬ 𝜒))
5	2, 4	ralprg 4654	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)))
6		ralnex 3087	. . . 4 ⊢ (∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑)
7		pm4.56 1001	. . . 4 ⊢ ((¬ 𝜓 ∧ ¬ 𝜒) ↔ ¬ (𝜓 ∨ 𝜒))
8	6, 7	bibi12i 341	. . 3 ⊢ ((∀𝑥 ∈ {𝐴, 𝐵} ¬ 𝜑 ↔ (¬ 𝜓 ∧ ¬ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
9		notbi 321	. . 3 ⊢ ((∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓 ∨ 𝜒)) ↔ (¬ ∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ¬ (𝜓 ∨ 𝜒)))
10	8, 9	sylbb2 240	. 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 208   ∧ wa 399   ∨ wo 858   = wceq 1559   ∈ wcel 2141  ∀wral 3075  ∃wrex 3085  {cpr 4583

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3076  df-rex 3086  df-v 3455  df-un 3909  df-sn 4582  df-pr 4584

This theorem is referenced by:  rextpg  4657  rexpr  4659  reurexprg  4662  fr2nr  5622  sgrp2nmndlem5  18949  nb3grprlem2  29528  nfrgr2v  30420  3vfriswmgrlem  30425  brfvrcld  44231  rnmptpr  45719  ldepspr  49059  zlmodzxzldeplem4  49089