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Theorem ssdifeq0 3305
 Description: A class is a subclass of itself subtracted from another iff it is the empty set. (Contributed by Steve Rodriguez, 20-Nov-2015.)
Assertion
Ref Expression
ssdifeq0 (𝐴 ⊆ (𝐵𝐴) ↔ 𝐴 = ∅)

Proof of Theorem ssdifeq0
StepHypRef Expression
1 inidm 3146 . . 3 (𝐴𝐴) = 𝐴
2 ssdifin0 3304 . . 3 (𝐴 ⊆ (𝐵𝐴) → (𝐴𝐴) = ∅)
31, 2syl5eqr 2086 . 2 (𝐴 ⊆ (𝐵𝐴) → 𝐴 = ∅)
4 0ss 3255 . . 3 ∅ ⊆ (𝐵 ∖ ∅)
5 id 19 . . . 4 (𝐴 = ∅ → 𝐴 = ∅)
6 difeq2 3056 . . . 4 (𝐴 = ∅ → (𝐵𝐴) = (𝐵 ∖ ∅))
75, 6sseq12d 2974 . . 3 (𝐴 = ∅ → (𝐴 ⊆ (𝐵𝐴) ↔ ∅ ⊆ (𝐵 ∖ ∅)))
84, 7mpbiri 157 . 2 (𝐴 = ∅ → 𝐴 ⊆ (𝐵𝐴))
93, 8impbii 117 1 (𝐴 ⊆ (𝐵𝐴) ↔ 𝐴 = ∅)
 Colors of variables: wff set class Syntax hints:   ↔ wb 98   = wceq 1243   ∖ cdif 2914   ∩ cin 2916   ⊆ wss 2917  ∅c0 3224 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rab 2315  df-v 2559  df-dif 2920  df-in 2924  df-ss 2931  df-nul 3225 This theorem is referenced by: (None)
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