Theorem csbcog

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

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

Proof

Step	Hyp	Ref	Expression
1		csbeq1	⊢ ( 𝑦 = 𝐴 → ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 ∘ 𝐶 ) = ⦋ 𝐴 / 𝑥 ⦌ ( 𝐵 ∘ 𝐶 ) )
2		csbeq1	⊢ ( 𝑦 = 𝐴 → ⦋ 𝑦 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐵 )
3		csbeq1	⊢ ( 𝑦 = 𝐴 → ⦋ 𝑦 / 𝑥 ⦌ 𝐶 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 )
4	2 3	coeq12d	⊢ ( 𝑦 = 𝐴 → ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) )
5	1 4	eqeq12d	⊢ ( 𝑦 = 𝐴 → ( ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 ∘ 𝐶 ) = ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) ↔ ⦋ 𝐴 / 𝑥 ⦌ ( 𝐵 ∘ 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) ) )
6		vex	⊢ 𝑦 ∈ V
7		nfcsb1v	⊢ Ⅎ 𝑥 ⦋ 𝑦 / 𝑥 ⦌ 𝐵
8		nfcsb1v	⊢ Ⅎ 𝑥 ⦋ 𝑦 / 𝑥 ⦌ 𝐶
9	7 8	nfco	⊢ Ⅎ 𝑥 ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝑦 / 𝑥 ⦌ 𝐶 )
10		csbeq1a	⊢ ( 𝑥 = 𝑦 → 𝐵 = ⦋ 𝑦 / 𝑥 ⦌ 𝐵 )
11		csbeq1a	⊢ ( 𝑥 = 𝑦 → 𝐶 = ⦋ 𝑦 / 𝑥 ⦌ 𝐶 )
12	10 11	coeq12d	⊢ ( 𝑥 = 𝑦 → ( 𝐵 ∘ 𝐶 ) = ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝑦 / 𝑥 ⦌ 𝐶 ) )
13	6 9 12	csbief	⊢ ⦋ 𝑦 / 𝑥 ⦌ ( 𝐵 ∘ 𝐶 ) = ( ⦋ 𝑦 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝑦 / 𝑥 ⦌ 𝐶 )
14	5 13	vtoclg	⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐵 ∘ 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∘ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) )