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Theorem difeq 30289
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 incom 4128 . . . . 5 (𝐵 ∩ (𝐴𝐵)) = ((𝐴𝐵) ∩ 𝐵)
2 disjdif 4379 . . . . 5 (𝐵 ∩ (𝐴𝐵)) = ∅
31, 2eqtr3i 2823 . . . 4 ((𝐴𝐵) ∩ 𝐵) = ∅
4 ineq1 4131 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∩ 𝐵) = (𝐶𝐵))
53, 4syl5reqr 2848 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = ∅)
6 undif1 4382 . . . 4 ((𝐴𝐵) ∪ 𝐵) = (𝐴𝐵)
7 uneq1 4083 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∪ 𝐵) = (𝐶𝐵))
86, 7syl5reqr 2848 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = (𝐴𝐵))
95, 8jca 515 . 2 ((𝐴𝐵) = 𝐶 → ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
10 simpl 486 . . . 4 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = ∅)
11 disj3 4361 . . . . 5 ((𝐶𝐵) = ∅ ↔ 𝐶 = (𝐶𝐵))
12 eqcom 2805 . . . . 5 (𝐶 = (𝐶𝐵) ↔ (𝐶𝐵) = 𝐶)
1311, 12bitri 278 . . . 4 ((𝐶𝐵) = ∅ ↔ (𝐶𝐵) = 𝐶)
1410, 13sylib 221 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = 𝐶)
15 difeq1 4043 . . . . . 6 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) ∖ 𝐵) = ((𝐴𝐵) ∖ 𝐵))
16 difun2 4387 . . . . . 6 ((𝐶𝐵) ∖ 𝐵) = (𝐶𝐵)
17 difun2 4387 . . . . . 6 ((𝐴𝐵) ∖ 𝐵) = (𝐴𝐵)
1815, 16, 173eqtr3g 2856 . . . . 5 ((𝐶𝐵) = (𝐴𝐵) → (𝐶𝐵) = (𝐴𝐵))
1918eqeq1d 2800 . . . 4 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
2019adantl 485 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
2114, 20mpbid 235 . 2 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐴𝐵) = 𝐶)
229, 21impbii 212 1 ((𝐴𝐵) = 𝐶 ↔ ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 399   = wceq 1538  cdif 3878  cun 3879  cin 3880  c0 4243
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-ral 3111  df-rab 3115  df-v 3443  df-dif 3884  df-un 3886  df-in 3888  df-ss 3898  df-nul 4244
This theorem is referenced by:  difioo  30531
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