Theorem fnund 6654

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

Hypotheses

Ref	Expression
fnund.1	⊢ (𝜑 → 𝐹 Fn 𝐴)
fnund.2	⊢ (𝜑 → 𝐺 Fn 𝐵)
fnund.3	⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)

Assertion

Ref	Expression
fnund	⊢ (𝜑 → (𝐹 ∪ 𝐺) Fn (𝐴 ∪ 𝐵))

Proof of Theorem fnund

Step	Hyp	Ref	Expression
1		fnund.1	. 2 ⊢ (𝜑 → 𝐹 Fn 𝐴)
2		fnund.2	. 2 ⊢ (𝜑 → 𝐺 Fn 𝐵)
3		fnund.3	. 2 ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)
4		fnun 6653	. 2 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺) Fn (𝐴 ∪ 𝐵))
5	1, 2, 3, 4	syl21anc 851	1 ⊢ (𝜑 → (𝐹 ∪ 𝐺) Fn (𝐴 ∪ 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∪ cun 3904   ∩ cin 3905  ∅c0 4286   Fn wfn 6535

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-12 2216  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-id 5558  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-fun 6542  df-fn 6543

This theorem is used by:  fnunop  6655  brwdom2  9538  sseqfn  34821  bnj927  35199  ofun  43039  tfsconcatfn  44098