Theorem fnund 6684

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 6683	. 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   ∩ cin 4092   ∪ cun 4152  ∅c0 4281   Fn wfn 6561

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-pr 5398

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-br 5109  df-opab 5174  df-id 5553  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-fun 6568  df-fn 6569

This theorem is used by:  fnunop  6685  brwdom2  9604  sseqfn  35133  bnj927  35511  ofun  43556  tfsconcatfn  44601