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