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Metamath Proof Explorer

Theorem fcompt

Description: Express composition of two functions as a maps-to applying both in sequence. (Contributed by Stefan O'Rear, 5-Oct-2014) (Proof shortened by Mario Carneiro, 27-Dec-2014)

Ref Expression
Assertion fcompt ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) → ( 𝐴 ∘ 𝐵 ) = ( 𝑥 ∈ 𝐶 ↦ ( 𝐴 ‘ ( 𝐵 ‘ 𝑥 ) ) ) )

Proof

Step Hyp Ref Expression
1 ffvelcdm ⊢ ( ( 𝐵 : 𝐶 ⟶ 𝐷 ∧ 𝑥 ∈ 𝐶 ) → ( 𝐵 ‘ 𝑥 ) ∈ 𝐷 )
2 1 adantll ⊢ ( ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) ∧ 𝑥 ∈ 𝐶 ) → ( 𝐵 ‘ 𝑥 ) ∈ 𝐷 )
3 ffn ⊢ ( 𝐵 : 𝐶 ⟶ 𝐷 → 𝐵 Fn 𝐶 )
4 3 adantl ⊢ ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) → 𝐵 Fn 𝐶 )
5 dffn5 ⊢ ( 𝐵 Fn 𝐶 ↔ 𝐵 = ( 𝑥 ∈ 𝐶 ↦ ( 𝐵 ‘ 𝑥 ) ) )
6 4 5 sylib ⊢ ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) → 𝐵 = ( 𝑥 ∈ 𝐶 ↦ ( 𝐵 ‘ 𝑥 ) ) )
7 ffn ⊢ ( 𝐴 : 𝐷 ⟶ 𝐸 → 𝐴 Fn 𝐷 )
8 7 adantr ⊢ ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) → 𝐴 Fn 𝐷 )
9 dffn5 ⊢ ( 𝐴 Fn 𝐷 ↔ 𝐴 = ( 𝑦 ∈ 𝐷 ↦ ( 𝐴 ‘ 𝑦 ) ) )
10 8 9 sylib ⊢ ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) → 𝐴 = ( 𝑦 ∈ 𝐷 ↦ ( 𝐴 ‘ 𝑦 ) ) )
11 fveq2 ⊢ ( 𝑦 = ( 𝐵 ‘ 𝑥 ) → ( 𝐴 ‘ 𝑦 ) = ( 𝐴 ‘ ( 𝐵 ‘ 𝑥 ) ) )
12 2 6 10 11 fmptco ⊢ ( ( 𝐴 : 𝐷 ⟶ 𝐸 ∧ 𝐵 : 𝐶 ⟶ 𝐷 ) → ( 𝐴 ∘ 𝐵 ) = ( 𝑥 ∈ 𝐶 ↦ ( 𝐴 ‘ ( 𝐵 ‘ 𝑥 ) ) ) )