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Theorem rexprg 3633
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 3588 . . . 4 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
21rexeqi 2670 . . 3 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑)
3 rexun 3307 . . 3 (∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑 ↔ (∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑))
42, 3bitri 183 . 2 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑))
5 ralprg.1 . . . . 5 (𝑥 = 𝐴 → (𝜑𝜓))
65rexsng 3622 . . . 4 (𝐴𝑉 → (∃𝑥 ∈ {𝐴}𝜑𝜓))
76orbi1d 786 . . 3 (𝐴𝑉 → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑)))
8 ralprg.2 . . . . 5 (𝑥 = 𝐵 → (𝜑𝜒))
98rexsng 3622 . . . 4 (𝐵𝑊 → (∃𝑥 ∈ {𝐵}𝜑𝜒))
109orbi2d 785 . . 3 (𝐵𝑊 → ((𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
117, 10sylan9bb 459 . 2 ((𝐴𝑉𝐵𝑊) → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
124, 11syl5bb 191 1 ((𝐴𝑉𝐵𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓𝜒)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104  wo 703   = wceq 1348  wcel 2141  wrex 2449  cun 3119  {csn 3581  {cpr 3582
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-rex 2454  df-v 2732  df-sbc 2956  df-un 3125  df-sn 3587  df-pr 3588
This theorem is referenced by:  rextpg  3635  rexpr  3637  minmax  11186  xrminmax  11221
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