Theorem fun2 6767

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 6766	. 2 ⊢ (((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶(𝐶 ∪ 𝐶))
2		unidm 4159	. . 3 ⊢ (𝐶 ∪ 𝐶) = 𝐶
3		feq3 6711	. . 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   ∩ cin 4092   ∪ cun 4152  ∅c0 4281  ⟶wf 6554

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 5252  ax-pr 5395

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 4839  df-br 5107  df-opab 5171  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-fun 6560  df-fn 6561  df-f 6562

This theorem is used by:  fun2d  6768  axlowdimlem5  29554  axlowdimlem7  29556  resf1o  33392  locfinref  34543  breprexplema  35329