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Theorem sbccomlem 2978
Description: Lemma for sbccom 2979. (Contributed by NM, 14-Nov-2005.) (Revised by Mario Carneiro, 18-Oct-2016.)
Assertion
Ref Expression
sbccomlem ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑[𝐵 / 𝑦][𝐴 / 𝑥]𝜑)
Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem sbccomlem
StepHypRef Expression
1 excom 1642 . . . 4 (∃𝑥𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑦𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
2 exdistr 1881 . . . 4 (∃𝑥𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
3 an12 550 . . . . . . 7 ((𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑦 = 𝐵 ∧ (𝑥 = 𝐴𝜑)))
43exbii 1584 . . . . . 6 (∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑥(𝑦 = 𝐵 ∧ (𝑥 = 𝐴𝜑)))
5 19.42v 1878 . . . . . 6 (∃𝑥(𝑦 = 𝐵 ∧ (𝑥 = 𝐴𝜑)) ↔ (𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
64, 5bitri 183 . . . . 5 (∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
76exbii 1584 . . . 4 (∃𝑦𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
81, 2, 73bitr3i 209 . . 3 (∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
9 sbc5 2927 . . 3 ([𝐴 / 𝑥]𝑦(𝑦 = 𝐵𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
10 sbc5 2927 . . 3 ([𝐵 / 𝑦]𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
118, 9, 103bitr4i 211 . 2 ([𝐴 / 𝑥]𝑦(𝑦 = 𝐵𝜑) ↔ [𝐵 / 𝑦]𝑥(𝑥 = 𝐴𝜑))
12 sbc5 2927 . . 3 ([𝐵 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝐵𝜑))
1312sbcbii 2963 . 2 ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑[𝐴 / 𝑥]𝑦(𝑦 = 𝐵𝜑))
14 sbc5 2927 . . 3 ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑))
1514sbcbii 2963 . 2 ([𝐵 / 𝑦][𝐴 / 𝑥]𝜑[𝐵 / 𝑦]𝑥(𝑥 = 𝐴𝜑))
1611, 13, 153bitr4i 211 1 ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑[𝐵 / 𝑦][𝐴 / 𝑥]𝜑)
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104   = wceq 1331  wex 1468  [wsbc 2904
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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-sbc 2905
This theorem is referenced by:  sbccom  2979
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