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Theorem difuncomp 30299
 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 incom 4177 . . . 4 (𝐶𝐴) = (𝐴𝐶)
2 sseqin2 4191 . . . . 5 (𝐴𝐶 ↔ (𝐶𝐴) = 𝐴)
32biimpi 218 . . . 4 (𝐴𝐶 → (𝐶𝐴) = 𝐴)
41, 3syl5reqr 2871 . . 3 (𝐴𝐶𝐴 = (𝐴𝐶))
54difeq1d 4097 . 2 (𝐴𝐶 → (𝐴𝐵) = ((𝐴𝐶) ∖ 𝐵))
6 difundi 4255 . . . 4 (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵))
7 dfss4 4234 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
87biimpi 218 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
98ineq1d 4187 . . . 4 (𝐴𝐶 → ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵)) = (𝐴 ∩ (𝐶𝐵)))
106, 9syl5eq 2868 . . 3 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶𝐵)))
11 indif2 4246 . . 3 (𝐴 ∩ (𝐶𝐵)) = ((𝐴𝐶) ∖ 𝐵)
1210, 11syl6eq 2872 . 2 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐴𝐶) ∖ 𝐵))
135, 12eqtr4d 2859 1 (𝐴𝐶 → (𝐴𝐵) = (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1533   ∖ cdif 3932   ∪ cun 3933   ∩ cin 3934   ⊆ wss 3935 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951 This theorem is referenced by:  ldgenpisyslem1  31417
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