Theorem f1un 6870

Description: The union of two one-to-one functions with disjoint domains and codomains. (Contributed by BTernaryTau, 3-Dec-2024.)

Assertion

Ref	Expression
f1un	⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(𝐵 ∪ 𝐷))

Proof of Theorem f1un

Step	Hyp	Ref	Expression
1		f1f 6803	. . . 4 ⊢ (𝐹:𝐴⟶1-1𝐵 → 𝐹:𝐴⟶𝐵)
2	1	frnd 6743	. . 3 ⊢ (𝐹:𝐴⟶1-1𝐵 → ran 𝐹 ⊆ 𝐵)
3		f1f 6803	. . . 4 ⊢ (𝐺:𝐶⟶1-1𝐷 → 𝐺:𝐶⟶𝐷)
4	3	frnd 6743	. . 3 ⊢ (𝐺:𝐶⟶1-1𝐷 → ran 𝐺 ⊆ 𝐷)
5		unss12 4189	. . 3 ⊢ ((ran 𝐹 ⊆ 𝐵 ∧ ran 𝐺 ⊆ 𝐷) → (ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷))
6	2, 4, 5	syl2an 608	. 2 ⊢ ((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) → (ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷))
7		f1f1orn 6861	. . . . 5 ⊢ (𝐹:𝐴⟶1-1𝐵 → 𝐹:𝐴⟶1-1ontoran 𝐹)
8		f1f1orn 6861	. . . . 5 ⊢ (𝐺:𝐶⟶1-1𝐷 → 𝐺:𝐶⟶1-1ontoran 𝐺)
9	7, 8	anim12i 625	. . . 4 ⊢ ((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) → (𝐹:𝐴⟶1-1ontoran 𝐹 ∧ 𝐺:𝐶⟶1-1ontoran 𝐺))
10		simprl 783	. . . . 5 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐴 ∩ 𝐶) = ∅)
11		ss2in 4144	. . . . . . . 8 ⊢ ((ran 𝐹 ⊆ 𝐵 ∧ ran 𝐺 ⊆ 𝐷) → (ran 𝐹 ∩ ran 𝐺) ⊆ (𝐵 ∩ 𝐷))
12	2, 4, 11	syl2an 608	. . . . . . 7 ⊢ ((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) → (ran 𝐹 ∩ ran 𝐺) ⊆ (𝐵 ∩ 𝐷))
13		sseq0 4357	. . . . . . 7 ⊢ (((ran 𝐹 ∩ ran 𝐺) ⊆ (𝐵 ∩ 𝐷) ∧ (𝐵 ∩ 𝐷) = ∅) → (ran 𝐹 ∩ ran 𝐺) = ∅)
14	12, 13	sylan 592	. . . . . 6 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ (𝐵 ∩ 𝐷) = ∅) → (ran 𝐹 ∩ ran 𝐺) = ∅)
15	14	adantrl 729	. . . . 5 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (ran 𝐹 ∩ ran 𝐺) = ∅)
16	10, 15	jca 521	. . . 4 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → ((𝐴 ∩ 𝐶) = ∅ ∧ (ran 𝐹 ∩ ran 𝐺) = ∅))
17		f1oun 6869	. . . 4 ⊢ (((𝐹:𝐴⟶1-1ontoran 𝐹 ∧ 𝐺:𝐶⟶1-1ontoran 𝐺) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (ran 𝐹 ∩ ran 𝐺) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1onto(ran 𝐹 ∪ ran 𝐺))
18	9, 16, 17	syl2an2r 698	. . 3 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1onto(ran 𝐹 ∪ ran 𝐺))
19		f1of1 6848	. . 3 ⊢ ((𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1onto(ran 𝐹 ∪ ran 𝐺) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(ran 𝐹 ∪ ran 𝐺))
20	18, 19	syl 18	. 2 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(ran 𝐹 ∪ ran 𝐺))
21		f1ss 6810	. . 3 ⊢ (((𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(ran 𝐹 ∪ ran 𝐺) ∧ (ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(𝐵 ∪ 𝐷))
22	21	ancoms 464	. 2 ⊢ (((ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷) ∧ (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(ran 𝐹 ∪ ran 𝐺)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(𝐵 ∪ 𝐷))
23	6, 20, 22	syl2an2r 698	1 ⊢ (((𝐹:𝐴⟶1-1𝐵 ∧ 𝐺:𝐶⟶1-1𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)⟶1-1(𝐵 ∪ 𝐷))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ⊆ wss 3900   ∩ cin 4092   ∪ cun 4152  ∅c0 4281  ran crn 5657  ⟶1-1wf1 6558  ⟶1-1ontowf1o 6560

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 5253  ax-pr 5396

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 5172  df-id 5551  df-xp 5662  df-rel 5663  df-cnv 5664  df-co 5665  df-dm 5666  df-rn 5667  df-fun 6563  df-fn 6564  df-f 6565  df-f1 6566  df-fo 6567  df-f1o 6568

This theorem is used by:  undom  9115