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Theorem uneqdifeq 4440
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 4105 . . . . 5 (𝐵𝐴) = (𝐴𝐵)
2 eqtr 2751 . . . . . . 7 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → (𝐵𝐴) = 𝐶)
32eqcomd 2737 . . . . . 6 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → 𝐶 = (𝐵𝐴))
4 difeq1 4066 . . . . . . 7 (𝐶 = (𝐵𝐴) → (𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴))
5 difun2 4428 . . . . . . 7 ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)
6 eqtr 2751 . . . . . . . 8 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → (𝐶𝐴) = (𝐵𝐴))
7 incom 4156 . . . . . . . . . . 11 (𝐴𝐵) = (𝐵𝐴)
87eqeq1i 2736 . . . . . . . . . 10 ((𝐴𝐵) = ∅ ↔ (𝐵𝐴) = ∅)
9 disj3 4401 . . . . . . . . . 10 ((𝐵𝐴) = ∅ ↔ 𝐵 = (𝐵𝐴))
108, 9bitri 275 . . . . . . . . 9 ((𝐴𝐵) = ∅ ↔ 𝐵 = (𝐵𝐴))
11 eqtr 2751 . . . . . . . . . . 11 (((𝐶𝐴) = (𝐵𝐴) ∧ (𝐵𝐴) = 𝐵) → (𝐶𝐴) = 𝐵)
1211expcom 413 . . . . . . . . . 10 ((𝐵𝐴) = 𝐵 → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1312eqcoms 2739 . . . . . . . . 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 4084 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐶)
23 sseq1 3955 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → ((𝐶𝐴) ⊆ 𝐶𝐵𝐶))
2422, 23mpbid 232 . . . . . . 7 ((𝐶𝐴) = 𝐵𝐵𝐶)
2524adantl 481 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐵𝐶)
2621, 25unssd 4139 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) ⊆ 𝐶)
27 eqimss 3988 . . . . . . 7 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐵)
28 ssundif 4435 . . . . . . 7 (𝐶 ⊆ (𝐴𝐵) ↔ (𝐶𝐴) ⊆ 𝐵)
2927, 28sylibr 234 . . . . . 6 ((𝐶𝐴) = 𝐵𝐶 ⊆ (𝐴𝐵))
3029adantl 481 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐶 ⊆ (𝐴𝐵))
3126, 30eqssd 3947 . . . 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 1541  cdif 3894  cun 3895  cin 3896  wss 3897  c0 4280
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ral 3048  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4281
This theorem is referenced by:  fzdifsuc  13484  hashbclem  14359  lecldbas  23134  conndisj  23331  ptuncnv  23722  ptunhmeo  23723  cldsubg  24026  icopnfcld  24682  iocmnfcld  24683  voliunlem1  25478  icombl  25492  ioombl  25493  uniioombllem4  25514  ismbf3d  25582  lhop  25948  symgcom  33052  f1resfz0f1d  35158  subfacp1lem3  35226  subfacp1lem5  35228  pconnconn  35275  cvmscld  35317
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