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 ( 𝜑𝐹 Fn 𝐴 )
fnund.2 ( 𝜑𝐺 Fn 𝐵 )
fnund.3 ( 𝜑 → ( 𝐴𝐵 ) = ∅ )
Assertion fnund ( 𝜑 → ( 𝐹𝐺 ) Fn ( 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 fnund.1 ( 𝜑𝐹 Fn 𝐴 )
2 fnund.2 ( 𝜑𝐺 Fn 𝐵 )
3 fnund.3 ( 𝜑 → ( 𝐴𝐵 ) = ∅ )
4 fnun ( ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐵 ) ∧ ( 𝐴𝐵 ) = ∅ ) → ( 𝐹𝐺 ) Fn ( 𝐴𝐵 ) )
5 1 2 3 4 syl21anc ( 𝜑 → ( 𝐹𝐺 ) Fn ( 𝐴𝐵 ) )