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Metamath Proof Explorer

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 ) )