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Theorem bj-sbeq 37906
Description: Distribute proper substitution through an equality relation. (See sbceqg 4369). (Contributed by BJ, 6-Oct-2018.)
Assertion
Ref Expression
bj-sbeq ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)

Proof of Theorem bj-sbeq
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 dfcleq 2753 . . . . 5 (𝐴 = 𝐵 ↔ ∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
21sbbii 2112 . . . 4 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ [𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
3 sbsbc 3742 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ [𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
4 sbcal 3797 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ∀𝑧[𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
52, 3, 43bitri 300 . . 3 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧[𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
6 sbcbig 3789 . . . . 5 (𝑦 ∈ V → ([𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵)))
76elv 3455 . . . 4 ([𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵))
87albii 1851 . . 3 (∀𝑧[𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ∀𝑧([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵))
9 sbcel2 4375 . . . . 5 ([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐴)
10 sbcel2 4375 . . . . 5 ([𝑦 / 𝑥]𝑧 ∈ 𝐵 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐵)
119, 10bibi12i 342 . . . 4 (([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵) ↔ (𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐴 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐵))
1211albii 1851 . . 3 (∀𝑧([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵) ↔ ∀𝑧(𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐴 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐵))
135, 8, 123bitri 300 . 2 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ∀𝑧(𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐴 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐵))
14 dfcleq 2753 . 2 (⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵 ↔ ∀𝑧(𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐴 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐵))
1513, 14bitr4i 281 1 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑦 / 𝑥⦌𝐵)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1567   = wceq 1569  [wsb 2098   ∈ wcel 2144  Vcvv 3450  [wsbc 3738  ⦋csb 3846
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-v 3452  df-sbc 3739  df-csb 3847  df-dif 4047  df-nul 4279
This theorem is used by: (None)
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