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Theorem difcom 4196
 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 3908 . . 3 (𝐵𝐶) = (𝐶𝐵)
21sseq2i 3779 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴 ⊆ (𝐶𝐵))
3 ssundif 4195 . 2 (𝐴 ⊆ (𝐵𝐶) ↔ (𝐴𝐵) ⊆ 𝐶)
4 ssundif 4195 . 2 (𝐴 ⊆ (𝐶𝐵) ↔ (𝐴𝐶) ⊆ 𝐵)
52, 3, 43bitr3i 290 1 ((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴𝐶) ⊆ 𝐵)
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 196   ∖ cdif 3720   ∪ cun 3721   ⊆ wss 3723 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-13 2408  ax-ext 2751 This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-clab 2758  df-cleq 2764  df-clel 2767  df-nfc 2902  df-v 3353  df-dif 3726  df-un 3728  df-in 3730  df-ss 3737 This theorem is referenced by:  pssdifcom1  4197  pssdifcom2  4198  isreg2  21402  restmetu  22595  conss1  39173  icccncfext  40615
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