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Theorem bj-sbeq 32540
Description: Distribute proper substitution through an equality relation. (See sbceqg 3956). (Contributed by BJ, 6-Oct-2018.)
Assertion
Ref Expression
bj-sbeq ([𝑦 / 𝑥]𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵)

Proof of Theorem bj-sbeq
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 dfcleq 2615 . . . . 5 (𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝐴𝑧𝐵))
21sbbii 1884 . . . 4 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ [𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵))
3 sbsbc 3421 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵) ↔ [𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵))
4 sbcal 3467 . . . 4 ([𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵) ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
52, 3, 43bitri 286 . . 3 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
6 vex 3189 . . . . 5 𝑦 ∈ V
7 sbcbig 3462 . . . . 5 (𝑦 ∈ V → ([𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵)))
86, 7ax-mp 5 . . . 4 ([𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵))
98albii 1744 . . 3 (∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵))
10 sbcel2 3961 . . . . 5 ([𝑦 / 𝑥]𝑧𝐴𝑧𝑦 / 𝑥𝐴)
11 sbcel2 3961 . . . . 5 ([𝑦 / 𝑥]𝑧𝐵𝑧𝑦 / 𝑥𝐵)
1210, 11bibi12i 329 . . . 4 (([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ (𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1312albii 1744 . . 3 (∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
145, 9, 133bitri 286 . 2 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
15 dfcleq 2615 . 2 (𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵 ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1614, 15bitr4i 267 1 ([𝑦 / 𝑥]𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵)
Colors of variables: wff setvar class
Syntax hints:  wb 196  wal 1478   = wceq 1480  [wsb 1877  wcel 1987  Vcvv 3186  [wsbc 3417  csb 3514
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-fal 1486  df-ex 1702  df-nf 1707  df-sb 1878  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-v 3188  df-sbc 3418  df-csb 3515  df-dif 3558  df-nul 3892
This theorem is referenced by: (None)
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