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Theorem bj-sbeq 37793
Description: Distribute proper substitution through an equality relation. (See sbceqg 4370). (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 2754 . . . . 5 (𝐴 = 𝐵 ↔ ∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
21sbbii 2113 . . . 4 ([𝑦 / 𝑥]𝐴 = 𝐵 ↔ [𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
3 sbsbc 3743 . . . 4 ([𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ [𝑦 / 𝑥]∀𝑧(𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵))
4 sbcal 3798 . . . 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 3790 . . . . 5 (𝑦 ∈ V → ([𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵)))
76elv 3456 . . . 4 ([𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵))
87albii 1852 . . 3 (∀𝑧[𝑦 / 𝑥](𝑧 ∈ 𝐴 ↔ 𝑧 ∈ 𝐵) ↔ ∀𝑧([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ [𝑦 / 𝑥]𝑧 ∈ 𝐵))
9 sbcel2 4376 . . . . 5 ([𝑦 / 𝑥]𝑧 ∈ 𝐴 ↔ 𝑧 ∈ ⦋𝑦 / 𝑥⦌𝐴)
10 sbcel2 4376 . . . . 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 1852 . . 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 2754 . 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 1568   = wceq 1570  [wsb 2099   ∈ wcel 2145  Vcvv 3451  [wsbc 3739  ⦋csb 3847
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-nul 4280
This theorem is used by: (None)
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