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Theorem ssofss 4077
Description: Condition for subset when A is already known to be a subset. (Contributed by SF, 13-Jan-2015.)
Assertion
Ref Expression
ssofss (A C → (A Bx C (x Ax B)))
Distinct variable groups:   x,A   x,B   x,C

Proof of Theorem ssofss
StepHypRef Expression
1 vex 2863 . . . . . . . 8 x V
21elcompl 3226 . . . . . . 7 (x C ↔ ¬ x C)
3 ssel 3268 . . . . . . . 8 (A C → (x Ax C))
43con3d 125 . . . . . . 7 (A C → (¬ x C → ¬ x A))
52, 4syl5bi 208 . . . . . 6 (A C → (x C → ¬ x A))
65imp 418 . . . . 5 ((A C x C) → ¬ x A)
76pm2.21d 98 . . . 4 ((A C x C) → (x Ax B))
87ralrimiva 2698 . . 3 (A Cx C(x Ax B))
98biantrud 493 . 2 (A C → (x C (x Ax B) ↔ (x C (x Ax B) x C(x Ax B))))
10 ralv 2873 . . . 4 (x V (x Ax B) ↔ x(x Ax B))
11 uncompl 4075 . . . . 5 (C ∪ ∼ C) = V
1211raleqi 2812 . . . 4 (x (C ∪ ∼ C)(x Ax B) ↔ x V (x Ax B))
13 dfss2 3263 . . . 4 (A Bx(x Ax B))
1410, 12, 133bitr4ri 269 . . 3 (A Bx (C ∪ ∼ C)(x Ax B))
15 ralunb 3445 . . 3 (x (C ∪ ∼ C)(x Ax B) ↔ (x C (x Ax B) x C(x Ax B)))
1614, 15bitri 240 . 2 (A B ↔ (x C (x Ax B) x C(x Ax B)))
179, 16syl6rbbr 255 1 (A C → (A Bx C (x Ax B)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 176   wa 358  wal 1540   wcel 1710  wral 2615  Vcvv 2860  ccompl 3206  cun 3208   wss 3258
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 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 2479  df-ral 2620  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-ss 3260
This theorem is referenced by:  ssofeq  4078  ssrelk  4212
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