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Theorem difcom 4488
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 4157 . . 3 (𝐵𝐶) = (𝐶𝐵)
21sseq2i 4012 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴 ⊆ (𝐶𝐵))
3 ssundif 4487 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴𝐵) ⊆ 𝐶)
4 ssundif 4487 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ (𝐴𝐶) ⊆ 𝐵)
52, 3, 43bitr3i 301 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴𝐶) ⊆ 𝐵)
Colors of variables: wff setvar class
Syntax hints:  wb 206  cdif 3947  cun 3948  wss 3950
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-v 3481  df-dif 3953  df-un 3955  df-ss 3967
This theorem is referenced by:  pssdifcom1  4489  pssdifcom2  4490  isreg2  23386  restmetu  24584  conss1  44468  icccncfext  45907
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