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Theorem uneqdifeq 4437
Description: Two ways to say that 𝐴 and 𝐵 partition 𝐶 (when 𝐴 and 𝐵 don't overlap and 𝐴 is a part of 𝐶). (Contributed by FL, 17-Nov-2008.) (Proof shortened by JJ, 14-Jul-2021.)
Assertion
Ref Expression
uneqdifeq ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 ↔ (𝐶𝐴) = 𝐵))

Proof of Theorem uneqdifeq
StepHypRef Expression
1 uncom 4128 . . . . 5 (𝐵𝐴) = (𝐴𝐵)
2 eqtr 2841 . . . . . . 7 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → (𝐵𝐴) = 𝐶)
32eqcomd 2827 . . . . . 6 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → 𝐶 = (𝐵𝐴))
4 difeq1 4091 . . . . . . 7 (𝐶 = (𝐵𝐴) → (𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴))
5 difun2 4428 . . . . . . 7 ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)
6 eqtr 2841 . . . . . . . 8 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → (𝐶𝐴) = (𝐵𝐴))
7 incom 4177 . . . . . . . . . . 11 (𝐴𝐵) = (𝐵𝐴)
87eqeq1i 2826 . . . . . . . . . 10 ((𝐴𝐵) = ∅ ↔ (𝐵𝐴) = ∅)
9 disj3 4402 . . . . . . . . . 10 ((𝐵𝐴) = ∅ ↔ 𝐵 = (𝐵𝐴))
108, 9bitri 277 . . . . . . . . 9 ((𝐴𝐵) = ∅ ↔ 𝐵 = (𝐵𝐴))
11 eqtr 2841 . . . . . . . . . . 11 (((𝐶𝐴) = (𝐵𝐴) ∧ (𝐵𝐴) = 𝐵) → (𝐶𝐴) = 𝐵)
1211expcom 416 . . . . . . . . . 10 ((𝐵𝐴) = 𝐵 → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1312eqcoms 2829 . . . . . . . . 9 (𝐵 = (𝐵𝐴) → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1410, 13sylbi 219 . . . . . . . 8 ((𝐴𝐵) = ∅ → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
156, 14syl5com 31 . . . . . . 7 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
164, 5, 15sylancl 588 . . . . . 6 (𝐶 = (𝐵𝐴) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
173, 16syl 17 . . . . 5 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
181, 17mpan 688 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
1918com12 32 . . 3 ((𝐴𝐵) = ∅ → ((𝐴𝐵) = 𝐶 → (𝐶𝐴) = 𝐵))
2019adantl 484 . 2 ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 → (𝐶𝐴) = 𝐵))
21 simpl 485 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐴𝐶)
22 difssd 4108 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐶)
23 sseq1 3991 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → ((𝐶𝐴) ⊆ 𝐶𝐵𝐶))
2422, 23mpbid 234 . . . . . . 7 ((𝐶𝐴) = 𝐵𝐵𝐶)
2524adantl 484 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐵𝐶)
2621, 25unssd 4161 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) ⊆ 𝐶)
27 eqimss 4022 . . . . . . 7 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐵)
28 ssundif 4432 . . . . . . 7 (𝐶 ⊆ (𝐴𝐵) ↔ (𝐶𝐴) ⊆ 𝐵)
2927, 28sylibr 236 . . . . . 6 ((𝐶𝐴) = 𝐵𝐶 ⊆ (𝐴𝐵))
3029adantl 484 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐶 ⊆ (𝐴𝐵))
3126, 30eqssd 3983 . . . 4 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) = 𝐶)
3231ex 415 . . 3 (𝐴𝐶 → ((𝐶𝐴) = 𝐵 → (𝐴𝐵) = 𝐶))
3332adantr 483 . 2 ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐶𝐴) = 𝐵 → (𝐴𝐵) = 𝐶))
3420, 33impbid 214 1 ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 ↔ (𝐶𝐴) = 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 398   = wceq 1533  cdif 3932  cun 3933  cin 3934  wss 3935  c0 4290
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291
This theorem is referenced by:  fzdifsuc  12961  hashbclem  13804  lecldbas  21821  conndisj  22018  ptuncnv  22409  ptunhmeo  22410  cldsubg  22713  icopnfcld  23370  iocmnfcld  23371  voliunlem1  24145  icombl  24159  ioombl  24160  uniioombllem4  24181  ismbf3d  24249  lhop  24607  symgcom  30722  f1resfz0f1d  32356  subfacp1lem3  32424  subfacp1lem5  32426  pconnconn  32473  cvmscld  32515
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