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Theorem ssdif0 3609
Description: Subclass expressed in terms of difference. Exercise 7 of [TakeutiZaring] p. 22. (Contributed by NM, 29-Apr-1994.)
Assertion
Ref Expression
ssdif0 (A B ↔ (A B) = )

Proof of Theorem ssdif0
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 iman 413 . . . 4 ((x Ax B) ↔ ¬ (x A ¬ x B))
2 eldif 3221 . . . 4 (x (A B) ↔ (x A ¬ x B))
31, 2xchbinxr 302 . . 3 ((x Ax B) ↔ ¬ x (A B))
43albii 1566 . 2 (x(x Ax B) ↔ x ¬ x (A B))
5 dfss2 3262 . 2 (A Bx(x Ax B))
6 eq0 3564 . 2 ((A B) = x ¬ x (A B))
74, 5, 63bitr4i 268 1 (A B ↔ (A B) = )
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 176   wa 358  wal 1540   = wceq 1642   wcel 1710   cdif 3206   wss 3257  c0 3550
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-ne 2518  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-ss 3259  df-nul 3551
This theorem is referenced by:  vdif0  3610  pssdifn0  3611  difid  3618  difin0  3623  sfinltfin  4535
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