Theorem csbcog 6300

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 3850	. . 3 ⊢ (𝑦 = 𝐴 → ⦋𝑦 / 𝑥⦌(𝐵 ∘ 𝐶) = ⦋𝐴 / 𝑥⦌(𝐵 ∘ 𝐶))
2		csbeq1 3850	. . . 4 ⊢ (𝑦 = 𝐴 → ⦋𝑦 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐵)
3		csbeq1 3850	. . . 4 ⊢ (𝑦 = 𝐴 → ⦋𝑦 / 𝑥⦌𝐶 = ⦋𝐴 / 𝑥⦌𝐶)
4	2, 3	coeq12d 5842	. . 3 ⊢ (𝑦 = 𝐴 → (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∘ ⦋𝐴 / 𝑥⦌𝐶))
5	1, 4	eqeq12d 2777	. 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 3455	. . 3 ⊢ 𝑦 ∈ V
7		nfcsb1v 3871	. . . 4 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐵
8		nfcsb1v 3871	. . . 4 ⊢ Ⅎ𝑥⦋𝑦 / 𝑥⦌𝐶
9	7, 8	nfco 5843	. . 3 ⊢ Ⅎ𝑥(⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶)
10		csbeq1a 3861	. . . 4 ⊢ (𝑥 = 𝑦 → 𝐵 = ⦋𝑦 / 𝑥⦌𝐵)
11		csbeq1a 3861	. . . 4 ⊢ (𝑥 = 𝑦 → 𝐶 = ⦋𝑦 / 𝑥⦌𝐶)
12	10, 11	coeq12d 5842	. . 3 ⊢ (𝑥 = 𝑦 → (𝐵 ∘ 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶))
13	6, 9, 12	csbief 3881	. 2 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 ∘ ⦋𝑦 / 𝑥⦌𝐶)
14	5, 13	vtoclg 3518	1 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝐵 ∘ 𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∘ ⦋𝐴 / 𝑥⦌𝐶))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2145  ⦋csb 3847   ∘ ccom 5655

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-rab 3414  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-co 5660

This theorem is used by:  sbcfung  6563  csbwrecsg  8336  brtrclfv2  44726