Theorem sbcbr 5204

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 5203	. 2 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3908	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3		csbconstg 3908	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
4	2, 3	breq12d 5162	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
5		br0 5198	. . . . 5 ⊢ ¬ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶
6		csbprc 4408	. . . . . 6 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝑅 = ∅)
7	6	breqd 5160	. . . . 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 5198	. . . . 5 ⊢ ¬ 𝐵∅𝐶
10	6	breqd 5160	. . . . 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 375	. . 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 2098  Vcvv 3461  [wsbc 3773  ⦋csb 3889  ∅c0 4322   class class class wbr 5149

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2166  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-nfc 2877  df-rab 3419  df-v 3463  df-sbc 3774  df-csb 3890  df-dif 3947  df-un 3949  df-ss 3961  df-nul 4323  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-br 5150

This theorem is referenced by:  csbcnv  5886