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