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Theorem difuncomp 32566
Description: Express a class difference using unions and class complements. (Contributed by Thierry Arnoux, 21-Jun-2020.)
Assertion
Ref Expression
difuncomp (𝐴𝐶 → (𝐴𝐵) = (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)))

Proof of Theorem difuncomp
StepHypRef Expression
1 sseqin2 4223 . . . . 5 (𝐴𝐶 ↔ (𝐶𝐴) = 𝐴)
21biimpi 216 . . . 4 (𝐴𝐶 → (𝐶𝐴) = 𝐴)
3 incom 4209 . . . 4 (𝐶𝐴) = (𝐴𝐶)
42, 3eqtr3di 2792 . . 3 (𝐴𝐶𝐴 = (𝐴𝐶))
54difeq1d 4125 . 2 (𝐴𝐶 → (𝐴𝐵) = ((𝐴𝐶) ∖ 𝐵))
6 difundi 4290 . . . 4 (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵))
7 dfss4 4269 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
87biimpi 216 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
98ineq1d 4219 . . . 4 (𝐴𝐶 → ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵)) = (𝐴 ∩ (𝐶𝐵)))
106, 9eqtrid 2789 . . 3 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶𝐵)))
11 indif2 4281 . . 3 (𝐴 ∩ (𝐶𝐵)) = ((𝐴𝐶) ∖ 𝐵)
1210, 11eqtrdi 2793 . 2 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐴𝐶) ∖ 𝐵))
135, 12eqtr4d 2780 1 (𝐴𝐶 → (𝐴𝐵) = (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1540  cdif 3948  cun 3949  cin 3950  wss 3951
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 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968
This theorem is referenced by:  ldgenpisyslem1  34164
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