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Theorem uneqdifeq 4493
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 4158 . . . . 5 (𝐵𝐴) = (𝐴𝐵)
2 eqtr 2760 . . . . . . 7 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → (𝐵𝐴) = 𝐶)
32eqcomd 2743 . . . . . 6 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → 𝐶 = (𝐵𝐴))
4 difeq1 4119 . . . . . . 7 (𝐶 = (𝐵𝐴) → (𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴))
5 difun2 4481 . . . . . . 7 ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)
6 eqtr 2760 . . . . . . . 8 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → (𝐶𝐴) = (𝐵𝐴))
7 incom 4209 . . . . . . . . . . 11 (𝐴𝐵) = (𝐵𝐴)
87eqeq1i 2742 . . . . . . . . . 10 ((𝐴𝐵) = ∅ ↔ (𝐵𝐴) = ∅)
9 disj3 4454 . . . . . . . . . 10 ((𝐵𝐴) = ∅ ↔ 𝐵 = (𝐵𝐴))
108, 9bitri 275 . . . . . . . . 9 ((𝐴𝐵) = ∅ ↔ 𝐵 = (𝐵𝐴))
11 eqtr 2760 . . . . . . . . . . 11 (((𝐶𝐴) = (𝐵𝐴) ∧ (𝐵𝐴) = 𝐵) → (𝐶𝐴) = 𝐵)
1211expcom 413 . . . . . . . . . 10 ((𝐵𝐴) = 𝐵 → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1312eqcoms 2745 . . . . . . . . 9 (𝐵 = (𝐵𝐴) → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1410, 13sylbi 217 . . . . . . . 8 ((𝐴𝐵) = ∅ → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
156, 14syl5com 31 . . . . . . 7 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
164, 5, 15sylancl 586 . . . . . 6 (𝐶 = (𝐵𝐴) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
173, 16syl 17 . . . . 5 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
181, 17mpan 690 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
1918com12 32 . . 3 ((𝐴𝐵) = ∅ → ((𝐴𝐵) = 𝐶 → (𝐶𝐴) = 𝐵))
2019adantl 481 . 2 ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 → (𝐶𝐴) = 𝐵))
21 simpl 482 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐴𝐶)
22 difssd 4137 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐶)
23 sseq1 4009 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → ((𝐶𝐴) ⊆ 𝐶𝐵𝐶))
2422, 23mpbid 232 . . . . . . 7 ((𝐶𝐴) = 𝐵𝐵𝐶)
2524adantl 481 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐵𝐶)
2621, 25unssd 4192 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) ⊆ 𝐶)
27 eqimss 4042 . . . . . . 7 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐵)
28 ssundif 4488 . . . . . . 7 (𝐶 ⊆ (𝐴𝐵) ↔ (𝐶𝐴) ⊆ 𝐵)
2927, 28sylibr 234 . . . . . 6 ((𝐶𝐴) = 𝐵𝐶 ⊆ (𝐴𝐵))
3029adantl 481 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐶 ⊆ (𝐴𝐵))
3126, 30eqssd 4001 . . . 4 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) = 𝐶)
3231ex 412 . . 3 (𝐴𝐶 → ((𝐶𝐴) = 𝐵 → (𝐴𝐵) = 𝐶))
3332adantr 480 . 2 ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐶𝐴) = 𝐵 → (𝐴𝐵) = 𝐶))
3420, 33impbid 212 1 ((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 ↔ (𝐶𝐴) = 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1540  cdif 3948  cun 3949  cin 3950  wss 3951  c0 4333
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334
This theorem is referenced by:  fzdifsuc  13624  hashbclem  14491  lecldbas  23227  conndisj  23424  ptuncnv  23815  ptunhmeo  23816  cldsubg  24119  icopnfcld  24788  iocmnfcld  24789  voliunlem1  25585  icombl  25599  ioombl  25600  uniioombllem4  25621  ismbf3d  25689  lhop  26055  symgcom  33103  f1resfz0f1d  35119  subfacp1lem3  35187  subfacp1lem5  35189  pconnconn  35236  cvmscld  35278
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