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Theorem rexprg 3760
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 3715 . . . 4 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
21rexeqi 2754 . . 3 (∃𝑥 ∈ {𝐴, 𝐵}𝜑 ↔ ∃𝑥 ∈ ({𝐴} ∪ {𝐵})𝜑)
3 rexun 3409 . . 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 3749 . . . 4 (𝐴𝑉 → (∃𝑥 ∈ {𝐴}𝜑𝜓))
76orbi1d 803 . . 3 (𝐴𝑉 → ((∃𝑥 ∈ {𝐴}𝜑 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑)))
8 ralprg.2 . . . . 5 (𝑥 = 𝐵 → (𝜑𝜒))
98rexsng 3749 . . . 4 (𝐵𝑊 → (∃𝑥 ∈ {𝐵}𝜑𝜒))
109orbi2d 802 . . 3 (𝐵𝑊 → ((𝜓 ∨ ∃𝑥 ∈ {𝐵}𝜑) ↔ (𝜓𝜒)))
117, 10sylan9bb 466 . 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 720   = wceq 1402  wcel 2209  wrex 2529  cun 3218  {csn 3708  {cpr 3709
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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-rex 2534  df-v 2823  df-sbc 3052  df-un 3224  df-sn 3714  df-pr 3715
This theorem is referenced by:  rextpg  3762  rexpr  3764  minmax  11979  xrminmax  12014
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