Theorem fresaunres1disj 5548

Description: From the union of two functions with disjoint domains, either component can be recovered by restriction. (Contributed by Mario Carneiro, 16-Feb-2015.) (Revised by Jim Kingdon, 18-May-2026.)

Assertion

Ref	Expression
fresaunres1disj	⊢ ((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐹 ∪ 𝐺) ↾ 𝐴) = 𝐹)

Proof of Theorem fresaunres1disj

Step	Hyp	Ref	Expression
1		fresaunres2disj 5547	. . 3 ⊢ ((𝐺:𝐵⟶𝐶 ∧ 𝐹:𝐴⟶𝐶 ∧ (𝐵 ∩ 𝐴) = ∅) → ((𝐺 ∪ 𝐹) ↾ 𝐴) = 𝐹)
2	1	3com12 1234	. 2 ⊢ ((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶 ∧ (𝐵 ∩ 𝐴) = ∅) → ((𝐺 ∪ 𝐹) ↾ 𝐴) = 𝐹)
3		incom 3413	. . . 4 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
4	3	eqeq1i 2242	. . 3 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ (𝐴 ∩ 𝐵) = ∅)
5	4	3anbi3i 1219	. 2 ⊢ ((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶 ∧ (𝐵 ∩ 𝐴) = ∅) ↔ (𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶 ∧ (𝐴 ∩ 𝐵) = ∅))
6		uncom 3365	. . . 4 ⊢ (𝐺 ∪ 𝐹) = (𝐹 ∪ 𝐺)
7	6	reseq1i 5036	. . 3 ⊢ ((𝐺 ∪ 𝐹) ↾ 𝐴) = ((𝐹 ∪ 𝐺) ↾ 𝐴)
8	7	eqeq1i 2242	. 2 ⊢ (((𝐺 ∪ 𝐹) ↾ 𝐴) = 𝐹 ↔ ((𝐹 ∪ 𝐺) ↾ 𝐴) = 𝐹)
9	2, 5, 8	3imtr3i 200	1 ⊢ ((𝐹:𝐴⟶𝐶 ∧ 𝐺:𝐵⟶𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐹 ∪ 𝐺) ↾ 𝐴) = 𝐹)

Syntax hints:   → wi 4   ∧ w3a 1005   = wceq 1398   ∪ cun 3211   ∩ cin 3212  ∅c0 3510   ↾ cres 4753  ⟶wf 5350

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-14 2208  ax-ext 2216  ax-sep 4230  ax-pow 4289  ax-pr 4324

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-v 2817  df-dif 3215  df-un 3217  df-in 3219  df-ss 3226  df-nul 3511  df-pw 3673  df-sn 3697  df-pr 3698  df-op 3700  df-br 4112  df-opab 4174  df-xp 4757  df-rel 4758  df-dm 4761  df-res 4763  df-fun 5356  df-fn 5357  df-f 5358

This theorem is referenced by:  mapunen  7106