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Theorem bj-sbeq 32088
 Description: Distribute proper substitution through an equality relation. (See sbceqg 3936). (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 2604 . . . . 5 (𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝐴𝑧𝐵))
21sbbii 1874 . . . 4 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ [𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵))
3 sbsbc 3406 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵) ↔ [𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵))
4 sbcal 3452 . . . 4 ([𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵) ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
52, 3, 43bitri 285 . . 3 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
6 vex 3176 . . . . 5 𝑦 ∈ V
7 sbcbig 3447 . . . . 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 1737 . . 3 (∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵))
10 sbcel2 3941 . . . . 5 ([𝑦 / 𝑥]𝑧𝐴𝑧𝑦 / 𝑥𝐴)
11 sbcel2 3941 . . . . 5 ([𝑦 / 𝑥]𝑧𝐵𝑧𝑦 / 𝑥𝐵)
1210, 11bibi12i 328 . . . 4 (([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ (𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1312albii 1737 . . 3 (∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
145, 9, 133bitri 285 . 2 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
15 dfcleq 2604 . 2 (𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵 ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1614, 15bitr4i 266 1 ([𝑦 / 𝑥]𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵)
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 195  ∀wal 1473   = wceq 1475  [wsb 1867   ∈ wcel 1977  Vcvv 3173  [wsbc 3402  ⦋csb 3499 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-fal 1481  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-sbc 3403  df-csb 3500  df-dif 3543  df-nul 3875 This theorem is referenced by: (None)
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