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Theorem difuncomp 31180
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 4162 . . . . 5 (𝐴𝐶 ↔ (𝐶𝐴) = 𝐴)
21biimpi 215 . . . 4 (𝐴𝐶 → (𝐶𝐴) = 𝐴)
3 incom 4148 . . . 4 (𝐶𝐴) = (𝐴𝐶)
42, 3eqtr3di 2791 . . 3 (𝐴𝐶𝐴 = (𝐴𝐶))
54difeq1d 4068 . 2 (𝐴𝐶 → (𝐴𝐵) = ((𝐴𝐶) ∖ 𝐵))
6 difundi 4226 . . . 4 (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵))
7 dfss4 4205 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
87biimpi 215 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
98ineq1d 4158 . . . 4 (𝐴𝐶 → ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵)) = (𝐴 ∩ (𝐶𝐵)))
106, 9eqtrid 2788 . . 3 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶𝐵)))
11 indif2 4217 . . 3 (𝐴 ∩ (𝐶𝐵)) = ((𝐴𝐶) ∖ 𝐵)
1210, 11eqtrdi 2792 . 2 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐴𝐶) ∖ 𝐵))
135, 12eqtr4d 2779 1 (𝐴𝐶 → (𝐴𝐵) = (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1540  cdif 3895  cun 3896  cin 3897  wss 3898
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2707
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2714  df-cleq 2728  df-clel 2814  df-rab 3404  df-v 3443  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915
This theorem is referenced by:  ldgenpisyslem1  32429
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