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Theorem rexprg 3698
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 3653 . . . 4 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
21rexeqi 2713 . . 3 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑)
3 rexun 3364 . . 3 (∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑 ↔ (∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑))
42, 3bitri 184 . 2 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑))
5 ralprg.1 . . . . 5 (𝑥 = 𝐴 → (𝜑𝜓))
65rexsng 3687 . . . 4 (𝐴𝑉 → (∃𝑥 ∈ {𝐴}𝜑𝜓))
76orbi1d 795 . . 3 (𝐴𝑉 → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑)))
8 ralprg.2 . . . . 5 (𝑥 = 𝐵 → (𝜑𝜒))
98rexsng 3687 . . . 4 (𝐵𝑊 → (∃𝑥 ∈ {𝐵}𝜑𝜒))
109orbi2d 794 . . 3 (𝐵𝑊 → ((𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
117, 10sylan9bb 462 . 2 ((𝐴𝑉𝐵𝑊) → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
124, 11bitrid 192 1 ((𝐴𝑉𝐵𝑊) → (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ (𝜓𝜒)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105  wo 712   = wceq 1375  wcel 2180  wrex 2489  cun 3175  {csn 3646  {cpr 3647
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191
This theorem depends on definitions:  df-bi 117  df-3an 985  df-tru 1378  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-rex 2494  df-v 2781  df-sbc 3009  df-un 3181  df-sn 3652  df-pr 3653
This theorem is referenced by:  rextpg  3700  rexpr  3702  minmax  11707  xrminmax  11742
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