Theorem fun2 6742

Description: The union of two functions with disjoint domains. (Contributed by Mario Carneiro, 12-Mar-2015.)

Assertion

Ref	Expression
fun2	⊢ (((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶𝐶)

Proof of Theorem fun2

Step	Hyp	Ref	Expression
1		fun 6741	. 2 ⊢ (((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶(𝐶 ∪ 𝐶))
2		unidm 4107	. . 3 ⊢ (𝐶 ∪ 𝐶) = 𝐶
3		feq3 6686	. . 3 ⊢ ((𝐶 ∪ 𝐶) = 𝐶 → ((𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶(𝐶 ∪ 𝐶) ↔ (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶𝐶))
4	2, 3	ax-mp 5	. 2 ⊢ ((𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶(𝐶 ∪ 𝐶) ↔ (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶𝐶)
5	1, 4	sylib 221	1 ⊢ (((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶𝐶)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∪ cun 3900   ∩ cin 3901  ∅c0 4282  ⟶wf 6533

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 2215  ax-ext 2734  ax-sep 5255  ax-pr 5402

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 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-id 5554  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-rn 5670  df-fun 6539  df-fn 6540  df-f 6541

This theorem is used by:  fun2d  6743  axlowdimlem5  29411  axlowdimlem7  29413  resf1o  33209  locfinref  34359  breprexplema  35146