Theorem sbcbr 5165

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 5164	. 2 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3884	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3		csbconstg 3884	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
4	2, 3	breq12d 5123	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
5		br0 5159	. . . . 5 ⊢ ¬ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶
6		csbprc 4375	. . . . . 6 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝑅 = ∅)
7	6	breqd 5121	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶))
8	5, 7	mtbiri 327	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
9		br0 5159	. . . . 5 ⊢ ¬ 𝐵∅𝐶
10	6	breqd 5121	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (𝐵⦋𝐴 / 𝑥⦌𝑅𝐶 ↔ 𝐵∅𝐶))
11	9, 10	mtbiri 327	. . . 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 275	1 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)

Syntax hints:  ¬ wn 3   ↔ wb 206   ∈ wcel 2109  Vcvv 3450  [wsbc 3756  ⦋csb 3865  ∅c0 4299   class class class wbr 5110

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-nfc 2879  df-rab 3409  df-v 3452  df-sbc 3757  df-csb 3866  df-dif 3920  df-un 3922  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-br 5111

This theorem is referenced by:  csbcnv  5850