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Theorem uneqdifeq 4459
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 4124 . . . . 5 (𝐵𝐴) = (𝐴𝐵)
2 eqtr 2750 . . . . . . 7 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → (𝐵𝐴) = 𝐶)
32eqcomd 2736 . . . . . 6 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → 𝐶 = (𝐵𝐴))
4 difeq1 4085 . . . . . . 7 (𝐶 = (𝐵𝐴) → (𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴))
5 difun2 4447 . . . . . . 7 ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)
6 eqtr 2750 . . . . . . . 8 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → (𝐶𝐴) = (𝐵𝐴))
7 incom 4175 . . . . . . . . . . 11 (𝐴𝐵) = (𝐵𝐴)
87eqeq1i 2735 . . . . . . . . . 10 ((𝐴𝐵) = ∅ ↔ (𝐵𝐴) = ∅)
9 disj3 4420 . . . . . . . . . 10 ((𝐵𝐴) = ∅ ↔ 𝐵 = (𝐵𝐴))
108, 9bitri 275 . . . . . . . . 9 ((𝐴𝐵) = ∅ ↔ 𝐵 = (𝐵𝐴))
11 eqtr 2750 . . . . . . . . . . 11 (((𝐶𝐴) = (𝐵𝐴) ∧ (𝐵𝐴) = 𝐵) → (𝐶𝐴) = 𝐵)
1211expcom 413 . . . . . . . . . 10 ((𝐵𝐴) = 𝐵 → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1312eqcoms 2738 . . . . . . . . 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 4103 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐶)
23 sseq1 3975 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → ((𝐶𝐴) ⊆ 𝐶𝐵𝐶))
2422, 23mpbid 232 . . . . . . 7 ((𝐶𝐴) = 𝐵𝐵𝐶)
2524adantl 481 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐵𝐶)
2621, 25unssd 4158 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) ⊆ 𝐶)
27 eqimss 4008 . . . . . . 7 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐵)
28 ssundif 4454 . . . . . . 7 (𝐶 ⊆ (𝐴𝐵) ↔ (𝐶𝐴) ⊆ 𝐵)
2927, 28sylibr 234 . . . . . 6 ((𝐶𝐴) = 𝐵𝐶 ⊆ (𝐴𝐵))
3029adantl 481 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐶 ⊆ (𝐴𝐵))
3126, 30eqssd 3967 . . . 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 3914  cun 3915  cin 3916  wss 3917  c0 4299
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 2008  ax-8 2111  ax-9 2119  ax-ext 2702
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-ral 3046  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-in 3924  df-ss 3934  df-nul 4300
This theorem is referenced by:  fzdifsuc  13552  hashbclem  14424  lecldbas  23113  conndisj  23310  ptuncnv  23701  ptunhmeo  23702  cldsubg  24005  icopnfcld  24662  iocmnfcld  24663  voliunlem1  25458  icombl  25472  ioombl  25473  uniioombllem4  25494  ismbf3d  25562  lhop  25928  symgcom  33047  f1resfz0f1d  35108  subfacp1lem3  35176  subfacp1lem5  35178  pconnconn  35225  cvmscld  35267
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