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Theorem difeq 32805
Description: Rewriting an equation with class difference, without using quantifiers. (Contributed by Thierry Arnoux, 24-Sep-2017.)
Assertion
Ref Expression
difeq ((𝐴𝐵) = 𝐶 ↔ ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))

Proof of Theorem difeq
StepHypRef Expression
1 ineq1 4174 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∩ 𝐵) = (𝐶𝐵))
2 disjdifr 4439 . . . 4 ((𝐴𝐵) ∩ 𝐵) = ∅
31, 2eqtr3di 2819 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = ∅)
4 uneq1 4123 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∪ 𝐵) = (𝐶𝐵))
5 undif1 4442 . . . 4 ((𝐴𝐵) ∪ 𝐵) = (𝐴𝐵)
64, 5eqtr3di 2819 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = (𝐴𝐵))
73, 6jca 520 . 2 ((𝐴𝐵) = 𝐶 → ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
8 disj3 4420 . . . . 5 ((𝐶𝐵) = ∅ ↔ 𝐶 = (𝐶𝐵))
9 eqcom 2776 . . . . 5 (𝐶 = (𝐶𝐵) ↔ (𝐶𝐵) = 𝐶)
108, 9bitri 278 . . . 4 ((𝐶𝐵) = ∅ ↔ (𝐶𝐵) = 𝐶)
1110birani 508 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = 𝐶)
12 difeq1 4082 . . . . . 6 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) ∖ 𝐵) = ((𝐴𝐵) ∖ 𝐵))
13 difun2 4447 . . . . . 6 ((𝐶𝐵) ∖ 𝐵) = (𝐶𝐵)
14 difun2 4447 . . . . . 6 ((𝐴𝐵) ∖ 𝐵) = (𝐴𝐵)
1512, 13, 143eqtr3g 2827 . . . . 5 ((𝐶𝐵) = (𝐴𝐵) → (𝐶𝐵) = (𝐴𝐵))
1615eqeq1d 2771 . . . 4 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1716adantl 486 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1811, 17mpbid 235 . 2 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐴𝐵) = 𝐶)
197, 18impbii 212 1 ((𝐴𝐵) = 𝐶 ↔ ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 400   = wceq 1567  cdif 3910  cun 3911  cin 3912  c0 4294
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295
This theorem is referenced by:  difioo  33068
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