Theorem sbcbr 5133

Description: Move substitution in and out of a binary relation. (Contributed by NM, 23-Aug-2018.)

Assertion

Ref	Expression
sbcbr	⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)

Distinct variable groups:   𝑥,𝐵   𝑥,𝐶

Allowed substitution hints:   𝐴(𝑥)   𝑅(𝑥)

Proof of Theorem sbcbr

Step	Hyp	Ref	Expression
1		sbcbr123 5132	. 2 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3855	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3		csbconstg 3855	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
4	2, 3	breq12d 5091	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
5		br0 5127	. . . . 5 ⊢ ¬ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶
6		csbprc 4345	. . . . . 6 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝑅 = ∅)
7	6	breqd 5089	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶))
8	5, 7	mtbiri 326	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
9		br0 5127	. . . . 5 ⊢ ¬ 𝐵∅𝐶
10	6	breqd 5089	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (𝐵⦋𝐴 / 𝑥⦌𝑅𝐶 ↔ 𝐵∅𝐶))
11	9, 10	mtbiri 326	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)
12	8, 11	2falsed 376	. . 3 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
13	4, 12	pm2.61i 182	. 2 ⊢ (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)
14	1, 13	bitri 274	1 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)

Syntax hints:  ¬ wn 3   ↔ wb 205   ∈ wcel 2109  Vcvv 3430  [wsbc 3719  ⦋csb 3836  ∅c0 4261   class class class wbr 5078

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-10 2140  ax-11 2157  ax-12 2174  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1544  df-fal 1554  df-ex 1786  df-nf 1790  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-nfc 2890  df-rab 3074  df-v 3432  df-sbc 3720  df-csb 3837  df-dif 3894  df-un 3896  df-nul 4262  df-if 4465  df-sn 4567  df-pr 4569  df-op 4573  df-br 5079

This theorem is referenced by:  csbcnv  5789