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Theorem rexprg 4206
Description: Convert a quantification over a pair to a disjunction. (Contributed by NM, 17-Sep-2011.) (Revised by Mario Carneiro, 23-Apr-2015.)
Hypotheses
Ref Expression
ralprg.1 (𝑥 = 𝐴 → (𝜑𝜓))
ralprg.2 (𝑥 = 𝐵 → (𝜑𝜒))
Assertion
Ref Expression
rexprg ((𝐴𝑉𝐵𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓𝜒)))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥   𝜒,𝑥
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)   𝑊(𝑥)

Proof of Theorem rexprg
StepHypRef Expression
1 df-pr 4151 . . . 4 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
21rexeqi 3132 . . 3 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑)
3 rexun 3771 . . 3 (∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑 ↔ (∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑))
42, 3bitri 264 . 2 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑))
5 ralprg.1 . . . . 5 (𝑥 = 𝐴 → (𝜑𝜓))
65rexsng 4190 . . . 4 (𝐴𝑉 → (∃𝑥 ∈ {𝐴}𝜑𝜓))
76orbi1d 738 . . 3 (𝐴𝑉 → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑)))
8 ralprg.2 . . . . 5 (𝑥 = 𝐵 → (𝜑𝜒))
98rexsng 4190 . . . 4 (𝐵𝑊 → (∃𝑥 ∈ {𝐵}𝜑𝜒))
109orbi2d 737 . . 3 (𝐵𝑊 → ((𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
117, 10sylan9bb 735 . 2 ((𝐴𝑉𝐵𝑊) → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
124, 11syl5bb 272 1 ((𝐴𝑉𝐵𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓𝜒)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wo 383  wa 384   = wceq 1480  wcel 1987  wrex 2908  cun 3553  {csn 4148  {cpr 4150
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1038  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-rex 2913  df-v 3188  df-sbc 3418  df-un 3560  df-sn 4149  df-pr 4151
This theorem is referenced by:  rextpg  4208  rexpr  4210  fr2nr  5052  sgrp2nmndlem5  17337  nb3grprlem2  26170  nfrgr2v  27000  3vfriswmgrlem  27005  brfvrcld  37461  rnmptpr  38829  ldepspr  41547  zlmodzxzldeplem4  41577
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