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Theorem difeq 32937
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 4166 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∩ 𝐵) = (𝐶𝐵))
2 disjdifr 4434 . . . 4 ((𝐴𝐵) ∩ 𝐵) = ∅
31, 2eqtr3di 2815 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = ∅)
4 uneq1 4115 . . . 4 ((𝐴𝐵) = 𝐶 → ((𝐴𝐵) ∪ 𝐵) = (𝐶𝐵))
5 undif1 4437 . . . 4 ((𝐴𝐵) ∪ 𝐵) = (𝐴𝐵)
64, 5eqtr3di 2815 . . 3 ((𝐴𝐵) = 𝐶 → (𝐶𝐵) = (𝐴𝐵))
73, 6jca 521 . 2 ((𝐴𝐵) = 𝐶 → ((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)))
8 disj3 4414 . . . . 5 ((𝐶𝐵) = ∅ ↔ 𝐶 = (𝐶𝐵))
9 eqcom 2772 . . . . 5 (𝐶 = (𝐶𝐵) ↔ (𝐶𝐵) = 𝐶)
108, 9bitri 278 . . . 4 ((𝐶𝐵) = ∅ ↔ (𝐶𝐵) = 𝐶)
1110birani 509 . . 3 (((𝐶𝐵) = ∅ ∧ (𝐶𝐵) = (𝐴𝐵)) → (𝐶𝐵) = 𝐶)
12 difeq1 4074 . . . . . 6 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) ∖ 𝐵) = ((𝐴𝐵) ∖ 𝐵))
13 difun2 4444 . . . . . 6 ((𝐶𝐵) ∖ 𝐵) = (𝐶𝐵)
14 difun2 4444 . . . . . 6 ((𝐴𝐵) ∖ 𝐵) = (𝐴𝐵)
1512, 13, 143eqtr3g 2823 . . . . 5 ((𝐶𝐵) = (𝐴𝐵) → (𝐶𝐵) = (𝐴𝐵))
1615eqeq1d 2767 . . . 4 ((𝐶𝐵) = (𝐴𝐵) → ((𝐶𝐵) = 𝐶 ↔ (𝐴𝐵) = 𝐶))
1716adantl 487 . . 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
This proof depends on syntax axioms:  wb 209  wa 401   = wceq 1570  cdif 3903  cun 3904  cin 3905  c0 4286
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287
This theorem is used by:  difioo  33199
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