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Theorem difuncomp 29206
 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 3788 . . . 4 (𝐶𝐴) = (𝐴𝐶)
2 sseqin2 3800 . . . . 5 (𝐴𝐶 ↔ (𝐶𝐴) = 𝐴)
32biimpi 206 . . . 4 (𝐴𝐶 → (𝐶𝐴) = 𝐴)
41, 3syl5reqr 2675 . . 3 (𝐴𝐶𝐴 = (𝐴𝐶))
54difeq1d 3710 . 2 (𝐴𝐶 → (𝐴𝐵) = ((𝐴𝐶) ∖ 𝐵))
6 difundi 3860 . . . 4 (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵))
7 dfss4 3841 . . . . . 6 (𝐴𝐶 ↔ (𝐶 ∖ (𝐶𝐴)) = 𝐴)
87biimpi 206 . . . . 5 (𝐴𝐶 → (𝐶 ∖ (𝐶𝐴)) = 𝐴)
98ineq1d 3796 . . . 4 (𝐴𝐶 → ((𝐶 ∖ (𝐶𝐴)) ∩ (𝐶𝐵)) = (𝐴 ∩ (𝐶𝐵)))
106, 9syl5eq 2672 . . 3 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶𝐵)))
11 indif2 3851 . . 3 (𝐴 ∩ (𝐶𝐵)) = ((𝐴𝐶) ∖ 𝐵)
1210, 11syl6eq 2676 . 2 (𝐴𝐶 → (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)) = ((𝐴𝐶) ∖ 𝐵))
135, 12eqtr4d 2663 1 (𝐴𝐶 → (𝐴𝐵) = (𝐶 ∖ ((𝐶𝐴) ∪ 𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1480   ∖ cdif 3557   ∪ cun 3558   ∩ cin 3559   ⊆ wss 3560 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1841  ax-6 1890  ax-7 1937  ax-9 2001  ax-10 2021  ax-11 2036  ax-12 2049  ax-13 2250  ax-ext 2606 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1883  df-clab 2613  df-cleq 2619  df-clel 2622  df-nfc 2756  df-ral 2917  df-rab 2921  df-v 3193  df-dif 3563  df-un 3565  df-in 3567  df-ss 3574 This theorem is referenced by:  ldgenpisyslem1  29999
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