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 → 2 nd ∘ F : A ⟶ C

Proof

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