Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  dfss6 Structured version   Visualization version   GIF version

Theorem dfss6 3626
 Description: Alternate definition of subclass relationship. (Contributed by RP, 16-Apr-2020.)
Assertion
Ref Expression
dfss6 (𝐴𝐵 ↔ ¬ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem dfss6
StepHypRef Expression
1 dfss2 3624 . . 3 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
2 notnotb 304 . . 3 (∀𝑥(𝑥𝐴𝑥𝐵) ↔ ¬ ¬ ∀𝑥(𝑥𝐴𝑥𝐵))
31, 2bitri 264 . 2 (𝐴𝐵 ↔ ¬ ¬ ∀𝑥(𝑥𝐴𝑥𝐵))
4 exanali 1826 . 2 (∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ¬ ∀𝑥(𝑥𝐴𝑥𝐵))
53, 4xchbinxr 324 1 (𝐴𝐵 ↔ ¬ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∧ wa 383  ∀wal 1521  ∃wex 1744   ∈ wcel 2030   ⊆ wss 3607 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-in 3614  df-ss 3621 This theorem is referenced by:  dfssr2  34389
 Copyright terms: Public domain W3C validator