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Theorem difeq 32504
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 4193 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∩ 𝐵) = (𝐶𝐵))
2 disjdifr 4453 . . . 4 ((𝐴𝐵) ∩ 𝐵) = ∅
31, 2eqtr3di 2786 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = ∅)
4 uneq1 4141 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∪ 𝐵) = (𝐶𝐵))
5 undif1 4456 . . . 4 ((𝐴𝐵) ∪ 𝐵) = (𝐴𝐵)
64, 5eqtr3di 2786 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = (𝐴𝐵))
73, 6jca 511 . 2 ((𝐴𝐵) = 𝐶 → ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
8 simpl 482 . . . 4 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = ∅)
9 disj3 4434 . . . . 5 ((𝐶𝐵) = ∅ ↔ 𝐶 = (𝐶𝐵))
10 eqcom 2743 . . . . 5 (𝐶 = (𝐶𝐵) ↔ (𝐶𝐵) = 𝐶)
119, 10bitri 275 . . . 4 ((𝐶𝐵) = ∅ ↔ (𝐶𝐵) = 𝐶)
128, 11sylib 218 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = 𝐶)
13 difeq1 4099 . . . . . 6 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) ∖ 𝐵) = ((𝐴𝐵) ∖ 𝐵))
14 difun2 4461 . . . . . 6 ((𝐶𝐵) ∖ 𝐵) = (𝐶𝐵)
15 difun2 4461 . . . . . 6 ((𝐴𝐵) ∖ 𝐵) = (𝐴𝐵)
1613, 14, 153eqtr3g 2794 . . . . 5 ((𝐶𝐵) = (𝐴𝐵) → (𝐶𝐵) = (𝐴𝐵))
1716eqeq1d 2738 . . . 4 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1817adantl 481 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1912, 18mpbid 232 . 2 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐴𝐵) = 𝐶)
207, 19impbii 209 1 ((𝐴𝐵) = 𝐶 ↔ ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395   = wceq 1540  cdif 3928  cun 3929  cin 3930  c0 4313
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 2708
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 2715  df-cleq 2728  df-clel 2810  df-ral 3053  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-in 3938  df-ss 3948  df-nul 4314
This theorem is referenced by:  difioo  32764
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