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Theorem difeq 31756
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 4206 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∩ 𝐵) = (𝐶𝐵))
2 disjdifr 4473 . . . 4 ((𝐴𝐵) ∩ 𝐵) = ∅
31, 2eqtr3di 2788 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = ∅)
4 uneq1 4157 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∪ 𝐵) = (𝐶𝐵))
5 undif1 4476 . . . 4 ((𝐴𝐵) ∪ 𝐵) = (𝐴𝐵)
64, 5eqtr3di 2788 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = (𝐴𝐵))
73, 6jca 513 . 2 ((𝐴𝐵) = 𝐶 → ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
8 simpl 484 . . . 4 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = ∅)
9 disj3 4454 . . . . 5 ((𝐶𝐵) = ∅ ↔ 𝐶 = (𝐶𝐵))
10 eqcom 2740 . . . . 5 (𝐶 = (𝐶𝐵) ↔ (𝐶𝐵) = 𝐶)
119, 10bitri 275 . . . 4 ((𝐶𝐵) = ∅ ↔ (𝐶𝐵) = 𝐶)
128, 11sylib 217 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = 𝐶)
13 difeq1 4116 . . . . . 6 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) ∖ 𝐵) = ((𝐴𝐵) ∖ 𝐵))
14 difun2 4481 . . . . . 6 ((𝐶𝐵) ∖ 𝐵) = (𝐶𝐵)
15 difun2 4481 . . . . . 6 ((𝐴𝐵) ∖ 𝐵) = (𝐴𝐵)
1613, 14, 153eqtr3g 2796 . . . . 5 ((𝐶𝐵) = (𝐴𝐵) → (𝐶𝐵) = (𝐴𝐵))
1716eqeq1d 2735 . . . 4 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1817adantl 483 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1912, 18mpbid 231 . 2 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐴𝐵) = 𝐶)
207, 19impbii 208 1 ((𝐴𝐵) = 𝐶 ↔ ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wa 397   = wceq 1542  cdif 3946  cun 3947  cin 3948  c0 4323
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3063  df-rab 3434  df-v 3477  df-dif 3952  df-un 3954  df-in 3956  df-ss 3966  df-nul 4324
This theorem is referenced by:  difioo  31993
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