Theorem fnund 6652

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 6651	. 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 3897   ∩ cin 3898  ∅c0 4279   Fn wfn 6532

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 2147  ax-9 2155  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-pr 5391

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 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-fun 6539  df-fn 6540

This theorem is used by:  fnunop  6653  brwdom2  9560  sseqfn  35015  bnj927  35393  ofun  43269  tfsconcatfn  44324