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Theorem difcom 4395
 Description: Swap the arguments of a class difference. (Contributed by NM, 29-Mar-2007.)
Assertion
Ref Expression
difcom ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴𝐶) ⊆ 𝐵)

Proof of Theorem difcom
StepHypRef Expression
1 uncom 4083 . . 3 (𝐵𝐶) = (𝐶𝐵)
21sseq2i 3947 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴 ⊆ (𝐶𝐵))
3 ssundif 4394 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴𝐵) ⊆ 𝐶)
4 ssundif 4394 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ (𝐴𝐶) ⊆ 𝐵)
52, 3, 43bitr3i 304 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴𝐶) ⊆ 𝐵)
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 209   ∖ cdif 3881   ∪ cun 3882   ⊆ wss 3884 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2114  ax-9 2122  ax-ext 2773 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2780  df-cleq 2794  df-clel 2873  df-v 3446  df-dif 3887  df-un 3889  df-in 3891  df-ss 3901 This theorem is referenced by:  pssdifcom1  4396  pssdifcom2  4397  isreg2  21986  restmetu  23181  conss1  41145  icccncfext  42526
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