Metamath Proof Explorer
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 ) ) |