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Theorem bj-sbeq 32896
Description: Distribute proper substitution through an equality relation. (See sbceqg 3984). (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 2616 . . . . 5 (𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝐴𝑧𝐵))
21sbbii 1887 . . . 4 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ [𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵))
3 sbsbc 3439 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵) ↔ [𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵))
4 sbcal 3485 . . . 4 ([𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵) ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
52, 3, 43bitri 286 . . 3 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
6 vex 3203 . . . . 5 𝑦 ∈ V
7 sbcbig 3480 . . . . 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 1747 . . 3 (∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵))
10 sbcel2 3989 . . . . 5 ([𝑦 / 𝑥]𝑧𝐴𝑧𝑦 / 𝑥𝐴)
11 sbcel2 3989 . . . . 5 ([𝑦 / 𝑥]𝑧𝐵𝑧𝑦 / 𝑥𝐵)
1210, 11bibi12i 329 . . . 4 (([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ (𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1312albii 1747 . . 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 2616 . 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 1481   = wceq 1483  [wsb 1880  wcel 1990  Vcvv 3200  [wsbc 3435  csb 3533
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-fal 1489  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-sbc 3436  df-csb 3534  df-dif 3577  df-nul 3916
This theorem is referenced by: (None)
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