Metamath Proof Explorer


Theorem fnund

Description: The union of two functions with disjoint domains, a deduction version. (Contributed by metakunt, 28-May-2024)

Ref Expression
Hypotheses fnund.1 φ F Fn A
fnund.2 φ G Fn B
fnund.3 φ A B =
Assertion fnund φ F G Fn A B

Proof

Step Hyp Ref Expression
1 fnund.1 φ F Fn A
2 fnund.2 φ G Fn B
3 fnund.3 φ A B =
4 fnun F Fn A G Fn B A B = F G Fn A B
5 1 2 3 4 syl21anc φ F G Fn A B