MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  ssindif0 Structured version   Visualization version   GIF version

Theorem ssindif0 4405
Description: Subclass expressed in terms of intersection with difference from the universal class. (Contributed by NM, 17-Sep-2003.)
Assertion
Ref Expression
ssindif0 (𝐴𝐵 ↔ (𝐴 ∩ (V ∖ 𝐵)) = ∅)

Proof of Theorem ssindif0
StepHypRef Expression
1 disj2 4399 . 2 ((𝐴 ∩ (V ∖ 𝐵)) = ∅ ↔ 𝐴 ⊆ (V ∖ (V ∖ 𝐵)))
2 ddif 4082 . . 3 (V ∖ (V ∖ 𝐵)) = 𝐵
32sseq2i 3952 . 2 (𝐴 ⊆ (V ∖ (V ∖ 𝐵)) ↔ 𝐴𝐵)
41, 3bitr2i 276 1 (𝐴𝐵 ↔ (𝐴 ∩ (V ∖ 𝐵)) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wb 206   = wceq 1542  Vcvv 3430  cdif 3887  cin 3889  wss 3890  c0 4274
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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-v 3432  df-dif 3893  df-in 3897  df-ss 3907  df-nul 4275
This theorem is referenced by:  setind  9659  setindregs  35290
  Copyright terms: Public domain W3C validator