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Theorem uneqdifeq 4516
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 4181 . . . . 5 (𝐵𝐴) = (𝐴𝐵)
2 eqtr 2763 . . . . . . 7 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → (𝐵𝐴) = 𝐶)
32eqcomd 2746 . . . . . 6 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → 𝐶 = (𝐵𝐴))
4 difeq1 4142 . . . . . . 7 (𝐶 = (𝐵𝐴) → (𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴))
5 difun2 4504 . . . . . . 7 ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)
6 eqtr 2763 . . . . . . . 8 (((𝐶𝐴) = ((𝐵𝐴) ∖ 𝐴) ∧ ((𝐵𝐴) ∖ 𝐴) = (𝐵𝐴)) → (𝐶𝐴) = (𝐵𝐴))
7 incom 4230 . . . . . . . . . . 11 (𝐴𝐵) = (𝐵𝐴)
87eqeq1i 2745 . . . . . . . . . 10 ((𝐴𝐵) = ∅ ↔ (𝐵𝐴) = ∅)
9 disj3 4477 . . . . . . . . . 10 ((𝐵𝐴) = ∅ ↔ 𝐵 = (𝐵𝐴))
108, 9bitri 275 . . . . . . . . 9 ((𝐴𝐵) = ∅ ↔ 𝐵 = (𝐵𝐴))
11 eqtr 2763 . . . . . . . . . . 11 (((𝐶𝐴) = (𝐵𝐴) ∧ (𝐵𝐴) = 𝐵) → (𝐶𝐴) = 𝐵)
1211expcom 413 . . . . . . . . . 10 ((𝐵𝐴) = 𝐵 → ((𝐶𝐴) = (𝐵𝐴) → (𝐶𝐴) = 𝐵))
1312eqcoms 2748 . . . . . . . . 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 585 . . . . . 6 (𝐶 = (𝐵𝐴) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
173, 16syl 17 . . . . 5 (((𝐵𝐴) = (𝐴𝐵) ∧ (𝐴𝐵) = 𝐶) → ((𝐴𝐵) = ∅ → (𝐶𝐴) = 𝐵))
181, 17mpan 689 . . . 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 4160 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐶)
23 sseq1 4034 . . . . . . . 8 ((𝐶𝐴) = 𝐵 → ((𝐶𝐴) ⊆ 𝐶𝐵𝐶))
2422, 23mpbid 232 . . . . . . 7 ((𝐶𝐴) = 𝐵𝐵𝐶)
2524adantl 481 . . . . . 6 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐵𝐶)
2621, 25unssd 4215 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → (𝐴𝐵) ⊆ 𝐶)
27 eqimss 4067 . . . . . . 7 ((𝐶𝐴) = 𝐵 → (𝐶𝐴) ⊆ 𝐵)
28 ssundif 4511 . . . . . . 7 (𝐶 ⊆ (𝐴𝐵) ↔ (𝐶𝐴) ⊆ 𝐵)
2927, 28sylibr 234 . . . . . 6 ((𝐶𝐴) = 𝐵𝐶 ⊆ (𝐴𝐵))
3029adantl 481 . . . . 5 ((𝐴𝐶 ∧ (𝐶𝐴) = 𝐵) → 𝐶 ⊆ (𝐴𝐵))
3126, 30eqssd 4026 . . . 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 1537  cdif 3973  cun 3974  cin 3975  wss 3976  c0 4352
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353
This theorem is referenced by:  fzdifsuc  13644  hashbclem  14501  lecldbas  23248  conndisj  23445  ptuncnv  23836  ptunhmeo  23837  cldsubg  24140  icopnfcld  24809  iocmnfcld  24810  voliunlem1  25604  icombl  25618  ioombl  25619  uniioombllem4  25640  ismbf3d  25708  lhop  26075  symgcom  33076  f1resfz0f1d  35081  subfacp1lem3  35150  subfacp1lem5  35152  pconnconn  35199  cvmscld  35241
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