Theorem fun2 6745

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 6744	. 2 ⊢ (((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶(𝐶 ∪ 𝐶))
2		unidm 4156	. . 3 ⊢ (𝐶 ∪ 𝐶) = 𝐶
3		feq3 6689	. . 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 1569   ∩ cin 4089   ∪ cun 4149  ∅c0 4278  ⟶wf 6534

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 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  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-rn 5661  df-fun 6540  df-fn 6541  df-f 6542

This theorem is used by:  fun2d  6746  axlowdimlem5  29529  axlowdimlem7  29531  resf1o  33368  locfinref  34519  breprexplema  35305