Metamath Proof Explorer


Theorem fun

Description: The union of two functions with disjoint domains. (Contributed by NM, 22-Sep-2004)

Ref Expression
Assertion fun ⊢ F : A ⟶ C ∧ G : B ⟶ D ∧ A ∩ B = ∅ → F ∪ G : A ∪ B ⟶ C ∪ D

Proof

Step Hyp Ref Expression
1 fnun ⊢ F Fn A ∧ G Fn B ∧ A ∩ B = ∅ → F ∪ G Fn A ∪ B
2 1 expcom ⊢ A ∩ B = ∅ → F Fn A ∧ G Fn B → F ∪ G Fn A ∪ B
3 rnun ⊢ ran ⁡ F ∪ G = ran ⁡ F ∪ ran ⁡ G
4 unss12 ⊢ ran ⁡ F ⊆ C ∧ ran ⁡ G ⊆ D → ran ⁡ F ∪ ran ⁡ G ⊆ C ∪ D
5 3 4 eqsstrid ⊢ ran ⁡ F ⊆ C ∧ ran ⁡ G ⊆ D → ran ⁡ F ∪ G ⊆ C ∪ D
6 2 5 anim12d1 ⊢ A ∩ B = ∅ → F Fn A ∧ G Fn B ∧ ran ⁡ F ⊆ C ∧ ran ⁡ G ⊆ D → F ∪ G Fn A ∪ B ∧ ran ⁡ F ∪ G ⊆ C ∪ D
7 df-f ⊢ F : A ⟶ C ↔ F Fn A ∧ ran ⁡ F ⊆ C
8 df-f ⊢ G : B ⟶ D ↔ G Fn B ∧ ran ⁡ G ⊆ D
9 7 8 anbi12i ⊢ F : A ⟶ C ∧ G : B ⟶ D ↔ F Fn A ∧ ran ⁡ F ⊆ C ∧ G Fn B ∧ ran ⁡ G ⊆ D
10 an4 ⊢ F Fn A ∧ ran ⁡ F ⊆ C ∧ G Fn B ∧ ran ⁡ G ⊆ D ↔ F Fn A ∧ G Fn B ∧ ran ⁡ F ⊆ C ∧ ran ⁡ G ⊆ D
11 9 10 bitri ⊢ F : A ⟶ C ∧ G : B ⟶ D ↔ F Fn A ∧ G Fn B ∧ ran ⁡ F ⊆ C ∧ ran ⁡ G ⊆ D
12 df-f ⊢ F ∪ G : A ∪ B ⟶ C ∪ D ↔ F ∪ G Fn A ∪ B ∧ ran ⁡ F ∪ G ⊆ C ∪ D
13 6 11 12 3imtr4g ⊢ A ∩ B = ∅ → F : A ⟶ C ∧ G : B ⟶ D → F ∪ G : A ∪ B ⟶ C ∪ D
14 13 impcom ⊢ F : A ⟶ C ∧ G : B ⟶ D ∧ A ∩ B = ∅ → F ∪ G : A ∪ B ⟶ C ∪ D