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Theorem sbccomlem 3541
Description: Lemma for sbccom 3542. (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 2082 . . . 4 (∃𝑥𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑦𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)))
2 exdistr 1922 . . . 4 (∃𝑥𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
3 an12 855 . . . . . . 7 ((𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑦 = 𝐵 ∧ (𝑥 = 𝐴𝜑)))
43exbii 1814 . . . . . 6 (∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑥(𝑦 = 𝐵 ∧ (𝑥 = 𝐴𝜑)))
5 19.42v 1921 . . . . . 6 (∃𝑥(𝑦 = 𝐵 ∧ (𝑥 = 𝐴𝜑)) ↔ (𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
64, 5bitri 264 . . . . 5 (∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ (𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
76exbii 1814 . . . 4 (∃𝑦𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
81, 2, 73bitr3i 290 . . 3 (∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
9 sbc5 3493 . . 3 ([𝐴 / 𝑥]𝑦(𝑦 = 𝐵𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵𝜑)))
10 sbc5 3493 . . 3 ([𝐵 / 𝑦]𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴𝜑)))
118, 9, 103bitr4i 292 . 2 ([𝐴 / 𝑥]𝑦(𝑦 = 𝐵𝜑) ↔ [𝐵 / 𝑦]𝑥(𝑥 = 𝐴𝜑))
12 sbc5 3493 . . 3 ([𝐵 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝐵𝜑))
1312sbcbii 3524 . 2 ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑[𝐴 / 𝑥]𝑦(𝑦 = 𝐵𝜑))
14 sbc5 3493 . . 3 ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑))
1514sbcbii 3524 . 2 ([𝐵 / 𝑦][𝐴 / 𝑥]𝜑[𝐵 / 𝑦]𝑥(𝑥 = 𝐴𝜑))
1611, 13, 153bitr4i 292 1 ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑[𝐵 / 𝑦][𝐴 / 𝑥]𝜑)
Colors of variables: wff setvar class
Syntax hints:  wb 196  wa 383   = wceq 1523  wex 1744  [wsbc 3468
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-v 3233  df-sbc 3469
This theorem is referenced by:  sbccom  3542
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