Metamath Proof Explorer


Theorem fmptcos

Description: Composition of two functions expressed as mapping abstractions. (Contributed by NM, 22-May-2006) (Revised by Mario Carneiro, 31-Aug-2015)

Ref Expression
Hypotheses fmptcof.1 ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 𝑅 ∈ 𝐵 )
fmptcof.2 ⊢ ( 𝜑 → 𝐹 = ( 𝑥 ∈ 𝐴 ↦ 𝑅 ) )
fmptcof.3 ⊢ ( 𝜑 → 𝐺 = ( 𝑦 ∈ 𝐵 ↦ 𝑆 ) )
Assertion fmptcos ( 𝜑 → ( 𝐺 ∘ 𝐹 ) = ( 𝑥 ∈ 𝐴 ↦ ⦋ 𝑅 / 𝑦 ⦌ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 fmptcof.1 ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 𝑅 ∈ 𝐵 )
2 fmptcof.2 ⊢ ( 𝜑 → 𝐹 = ( 𝑥 ∈ 𝐴 ↦ 𝑅 ) )
3 fmptcof.3 ⊢ ( 𝜑 → 𝐺 = ( 𝑦 ∈ 𝐵 ↦ 𝑆 ) )
4 nfcv ⊢ Ⅎ 𝑧 𝑆
5 nfcsb1v ⊢ Ⅎ 𝑦 ⦋ 𝑧 / 𝑦 ⦌ 𝑆
6 csbeq1a ⊢ ( 𝑦 = 𝑧 → 𝑆 = ⦋ 𝑧 / 𝑦 ⦌ 𝑆 )
7 4 5 6 cbvmpt ⊢ ( 𝑦 ∈ 𝐵 ↦ 𝑆 ) = ( 𝑧 ∈ 𝐵 ↦ ⦋ 𝑧 / 𝑦 ⦌ 𝑆 )
8 3 7 eqtrdi ⊢ ( 𝜑 → 𝐺 = ( 𝑧 ∈ 𝐵 ↦ ⦋ 𝑧 / 𝑦 ⦌ 𝑆 ) )
9 csbeq1 ⊢ ( 𝑧 = 𝑅 → ⦋ 𝑧 / 𝑦 ⦌ 𝑆 = ⦋ 𝑅 / 𝑦 ⦌ 𝑆 )
10 1 2 8 9 fmptcof ⊢ ( 𝜑 → ( 𝐺 ∘ 𝐹 ) = ( 𝑥 ∈ 𝐴 ↦ ⦋ 𝑅 / 𝑦 ⦌ 𝑆 ) )