Theorem csbcog 6307

Description: Distribute proper substitution through a composition of relations. (Contributed by RP, 28-Jun-2020.)

Assertion

Ref	Expression
csbcog	⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∘ ⦋𝐴 / 𝑥⦌𝐶))

Proof of Theorem csbcog

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		csbeq1 3854	. . 3 ⊢ (𝑦 = 𝐴 → ⦋𝑦 / 𝑥⦌(𝐵 ∘ 𝐶) = ⦋𝐴 / 𝑥⦌(𝐵 ∘ 𝐶))
2		csbeq1 3854	. . . 4 ⊢ (𝑦 = 𝐴 → ⦋𝑦 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐵)
3		csbeq1 3854	. . . 4 ⊢ (𝑦 = 𝐴 → ⦋𝑦 / 𝑥⦌𝐶 = ⦋𝐴 / 𝑥⦌𝐶)
4	2, 3	coeq12d 5847	. . 3 ⊢ (𝑦 = 𝐴 → (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∘ ⦋𝐴 / 𝑥⦌𝐶))
5	1, 4	eqeq12d 2779	. 2 ⊢ (𝑦 = 𝐴 → (⦋𝑦 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶) ↔ ⦋𝐴 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∘ ⦋𝐴 / 𝑥⦌𝐶)))
6		vex 3458	. . 3 ⊢ 𝑦 ∈ V
7		nfcsb1v 3875	. . . 4 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐵
8		nfcsb1v 3875	. . . 4 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐶
9	7, 8	nfco 5848	. . 3 ⊢ Ⅎ𝑥(⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶)
10		csbeq1a 3865	. . . 4 ⊢ (𝑥 = 𝑦 → 𝐵 = ⦋𝑦 / 𝑥⦌𝐵)
11		csbeq1a 3865	. . . 4 ⊢ (𝑥 = 𝑦 → 𝐶 = ⦋𝑦 / 𝑥⦌𝐶)
12	10, 11	coeq12d 5847	. . 3 ⊢ (𝑥 = 𝑦 → (𝐵 ∘ 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶))
13	6, 9, 12	csbief 3885	. 2 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶)
14	5, 13	vtoclg 3521	1 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∘ ⦋𝐴 / 𝑥⦌𝐶))

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ∈ wcel 2146  ⦋csb 3851   ∘ ccom 5659

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2196  ax-12 2215  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5171  df-co 5664

This theorem is used by:  sbcfung  6572  csbwrecsg  8350  brtrclfv2  44870