Theorem fnund 6657

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 6656	. 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 1569   ∩ cin 4088   ∪ cun 4148  ∅c0 4277   Fn wfn 6536

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-id 5545  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-fun 6543  df-fn 6544

This theorem is used by:  fnunop  6658  brwdom2  9572  sseqfn  35070  bnj927  35448  ofun  43396  tfsconcatfn  44447