Metamath Proof Explorer


Theorem 2ndcof

Description: Composition of the second member function with another function. (Contributed by FL, 15-Oct-2012)

Ref Expression
Assertion 2ndcof ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ( 2nd𝐹 ) : 𝐴𝐶 )

Proof

Step Hyp Ref Expression
1 fo2nd 2nd : V –onto→ V
2 fofn ( 2nd : V –onto→ V → 2nd Fn V )
3 1 2 ax-mp 2nd Fn V
4 ffn ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → 𝐹 Fn 𝐴 )
5 dffn2 ( 𝐹 Fn 𝐴𝐹 : 𝐴 ⟶ V )
6 4 5 sylib ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → 𝐹 : 𝐴 ⟶ V )
7 fnfco ( ( 2nd Fn V ∧ 𝐹 : 𝐴 ⟶ V ) → ( 2nd𝐹 ) Fn 𝐴 )
8 3 6 7 sylancr ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ( 2nd𝐹 ) Fn 𝐴 )
9 rnco ran ( 2nd𝐹 ) = ran ( 2nd ↾ ran 𝐹 )
10 frn ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ran 𝐹 ⊆ ( 𝐵 × 𝐶 ) )
11 ssres2 ( ran 𝐹 ⊆ ( 𝐵 × 𝐶 ) → ( 2nd ↾ ran 𝐹 ) ⊆ ( 2nd ↾ ( 𝐵 × 𝐶 ) ) )
12 rnss ( ( 2nd ↾ ran 𝐹 ) ⊆ ( 2nd ↾ ( 𝐵 × 𝐶 ) ) → ran ( 2nd ↾ ran 𝐹 ) ⊆ ran ( 2nd ↾ ( 𝐵 × 𝐶 ) ) )
13 10 11 12 3syl ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ran ( 2nd ↾ ran 𝐹 ) ⊆ ran ( 2nd ↾ ( 𝐵 × 𝐶 ) ) )
14 f2ndres ( 2nd ↾ ( 𝐵 × 𝐶 ) ) : ( 𝐵 × 𝐶 ) ⟶ 𝐶
15 frn ( ( 2nd ↾ ( 𝐵 × 𝐶 ) ) : ( 𝐵 × 𝐶 ) ⟶ 𝐶 → ran ( 2nd ↾ ( 𝐵 × 𝐶 ) ) ⊆ 𝐶 )
16 14 15 ax-mp ran ( 2nd ↾ ( 𝐵 × 𝐶 ) ) ⊆ 𝐶
17 13 16 sstrdi ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ran ( 2nd ↾ ran 𝐹 ) ⊆ 𝐶 )
18 9 17 eqsstrid ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ran ( 2nd𝐹 ) ⊆ 𝐶 )
19 df-f ( ( 2nd𝐹 ) : 𝐴𝐶 ↔ ( ( 2nd𝐹 ) Fn 𝐴 ∧ ran ( 2nd𝐹 ) ⊆ 𝐶 ) )
20 8 18 19 sylanbrc ( 𝐹 : 𝐴 ⟶ ( 𝐵 × 𝐶 ) → ( 2nd𝐹 ) : 𝐴𝐶 )