Theorem sbcbr 5155

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 5154	. 2 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3871	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3		csbconstg 3871	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
4	2, 3	breq12d 5113	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
5		br0 5149	. . . . 5 ⊢ ¬ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶
6		csbprc 4363	. . . . . 6 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝑅 = ∅)
7	6	breqd 5111	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶))
8	5, 7	mtbiri 329	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
9		br0 5149	. . . . 5 ⊢ ¬ 𝐵∅𝐶
10	6	breqd 5111	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (𝐵⦋𝐴 / 𝑥⦌𝑅𝐶 ↔ 𝐵∅𝐶))
11	9, 10	mtbiri 329	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)
12	8, 11	2falsed 378	. . 3 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
13	4, 12	pm2.61i 183	. 2 ⊢ (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)
14	1, 13	bitri 277	1 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)

Syntax hints:  ¬ wn 3   ↔ wb 208   ∈ wcel 2142  Vcvv 3454  [wsbc 3744  ⦋csb 3852  ∅c0 4285   class class class wbr 5100

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-nf 1804  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-rab 3415  df-v 3456  df-sbc 3745  df-csb 3853  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101

This theorem is referenced by:  csbcnv  5858  csbcnvOLD  5859