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Theorem bj-sbeq 35013
Description: Distribute proper substitution through an equality relation. (See sbceqg 4340). (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 2731 . . . . 5 (𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝐴𝑧𝐵))
21sbbii 2080 . . . 4 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ [𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵))
3 sbsbc 3715 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧𝐴𝑧𝐵) ↔ [𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵))
4 sbcal 3776 . . . 4 ([𝑦 / 𝑥]𝑧(𝑧𝐴𝑧𝐵) ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
52, 3, 43bitri 296 . . 3 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵))
6 sbcbig 3765 . . . . 5 (𝑦 ∈ V → ([𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵)))
76elv 3428 . . . 4 ([𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵))
87albii 1823 . . 3 (∀𝑧[𝑦 / 𝑥](𝑧𝐴𝑧𝐵) ↔ ∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵))
9 sbcel2 4346 . . . . 5 ([𝑦 / 𝑥]𝑧𝐴𝑧𝑦 / 𝑥𝐴)
10 sbcel2 4346 . . . . 5 ([𝑦 / 𝑥]𝑧𝐵𝑧𝑦 / 𝑥𝐵)
119, 10bibi12i 339 . . . 4 (([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ (𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1211albii 1823 . . 3 (∀𝑧([𝑦 / 𝑥]𝑧𝐴[𝑦 / 𝑥]𝑧𝐵) ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
135, 8, 123bitri 296 . 2 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
14 dfcleq 2731 . 2 (𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵 ↔ ∀𝑧(𝑧𝑦 / 𝑥𝐴𝑧𝑦 / 𝑥𝐵))
1513, 14bitr4i 277 1 ([𝑦 / 𝑥]𝐴 = 𝐵𝑦 / 𝑥𝐴 = 𝑦 / 𝑥𝐵)
Colors of variables: wff setvar class
Syntax hints:  wb 205  wal 1537   = wceq 1539  [wsb 2068  wcel 2108  Vcvv 3422  [wsbc 3711  csb 3828
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-v 3424  df-sbc 3712  df-csb 3829  df-dif 3886  df-nul 4254
This theorem is referenced by: (None)
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