Theorem sbcbr 4898

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 4897	. 2 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3741	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3		csbconstg 3741	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
4	2, 3	breq12d 4856	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
5		br0 4892	. . . . 5 ⊢ ¬ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶
6		csbprc 4176	. . . . . 6 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝑅 = ∅)
7	6	breqd 4854	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵∅⦋𝐴 / 𝑥⦌𝐶))
8	5, 7	mtbiri 319	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ ⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶)
9		br0 4892	. . . . 5 ⊢ ¬ 𝐵∅𝐶
10	6	breqd 4854	. . . . 5 ⊢ (¬ 𝐴 ∈ V → (𝐵⦋𝐴 / 𝑥⦌𝑅𝐶 ↔ 𝐵∅𝐶))
11	9, 10	mtbiri 319	. . . 4 ⊢ (¬ 𝐴 ∈ V → ¬ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)
12	8, 11	2falsed 368	. . 3 ⊢ (¬ 𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶))
13	4, 12	pm2.61i 177	. 2 ⊢ (⦋𝐴 / 𝑥⦌𝐵⦋𝐴 / 𝑥⦌𝑅⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)
14	1, 13	bitri 267	1 ⊢ ([𝐴 / 𝑥]𝐵𝑅𝐶 ↔ 𝐵⦋𝐴 / 𝑥⦌𝑅𝐶)

Syntax hints:  ¬ wn 3   ↔ wb 198   ∈ wcel 2157  Vcvv 3385  [wsbc 3633  ⦋csb 3728  ∅c0 4115   class class class wbr 4843

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-3an 1110  df-tru 1657  df-fal 1667  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-rab 3098  df-v 3387  df-sbc 3634  df-csb 3729  df-dif 3772  df-un 3774  df-in 3776  df-ss 3783  df-nul 4116  df-if 4278  df-sn 4369  df-pr 4371  df-op 4375  df-br 4844

This theorem is referenced by:  csbcnv  5509