Theorem fun2 6737

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 6736	. 2 ⊢ (((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐵)⟶(𝐶 ∪ 𝐶))
2		unidm 4104	. . 3 ⊢ (𝐶 ∪ 𝐶) = 𝐶
3		feq3 6681	. . 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 3897   ∩ cin 3898  ∅c0 4279  ⟶wf 6527

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-rn 5662  df-fun 6533  df-fn 6534  df-f 6535

This theorem is used by:  fun2d  6738  axlowdimlem5  29506  axlowdimlem7  29508  resf1o  33304  locfinref  34455  breprexplema  35242