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Metamath Proof Explorer

Theorem fnun

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

Ref Expression
Assertion fnun ( ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 df-fn ⊢ ( 𝐹 Fn 𝐴 ↔ ( Fun 𝐹 ∧ dom 𝐹 = 𝐴 ) )
2 df-fn ⊢ ( 𝐺 Fn 𝐵 ↔ ( Fun 𝐺 ∧ dom 𝐺 = 𝐵 ) )
3 ineq12 ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( dom 𝐹 ∩ dom 𝐺 ) = ( 𝐴 ∩ 𝐵 ) )
4 3 eqeq1d ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( ( dom 𝐹 ∩ dom 𝐺 ) = ∅ ↔ ( 𝐴 ∩ 𝐵 ) = ∅ ) )
5 4 anbi2d ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( dom 𝐹 ∩ dom 𝐺 ) = ∅ ) ↔ ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) ) )
6 funun ⊢ ( ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( dom 𝐹 ∩ dom 𝐺 ) = ∅ ) → Fun ( 𝐹 ∪ 𝐺 ) )
7 5 6 biimtrrdi ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → Fun ( 𝐹 ∪ 𝐺 ) ) )
8 dmun ⊢ dom ( 𝐹 ∪ 𝐺 ) = ( dom 𝐹 ∪ dom 𝐺 )
9 uneq12 ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( dom 𝐹 ∪ dom 𝐺 ) = ( 𝐴 ∪ 𝐵 ) )
10 8 9 eqtrid ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → dom ( 𝐹 ∪ 𝐺 ) = ( 𝐴 ∪ 𝐵 ) )
11 7 10 jctird ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( Fun ( 𝐹 ∪ 𝐺 ) ∧ dom ( 𝐹 ∪ 𝐺 ) = ( 𝐴 ∪ 𝐵 ) ) ) )
12 df-fn ⊢ ( ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) ↔ ( Fun ( 𝐹 ∪ 𝐺 ) ∧ dom ( 𝐹 ∪ 𝐺 ) = ( 𝐴 ∪ 𝐵 ) ) )
13 11 12 imbitrrdi ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) ) )
14 13 expd ⊢ ( ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) → ( ( Fun 𝐹 ∧ Fun 𝐺 ) → ( ( 𝐴 ∩ 𝐵 ) = ∅ → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) ) ) )
15 14 impcom ⊢ ( ( ( Fun 𝐹 ∧ Fun 𝐺 ) ∧ ( dom 𝐹 = 𝐴 ∧ dom 𝐺 = 𝐵 ) ) → ( ( 𝐴 ∩ 𝐵 ) = ∅ → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) ) )
16 15 an4s ⊢ ( ( ( Fun 𝐹 ∧ dom 𝐹 = 𝐴 ) ∧ ( Fun 𝐺 ∧ dom 𝐺 = 𝐵 ) ) → ( ( 𝐴 ∩ 𝐵 ) = ∅ → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) ) )
17 1 2 16 syl2anb ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵 ) → ( ( 𝐴 ∩ 𝐵 ) = ∅ → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) ) )
18 17 imp ⊢ ( ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( 𝐹 ∪ 𝐺 ) Fn ( 𝐴 ∪ 𝐵 ) )