MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  pssdifcom2 Structured version   Visualization version   GIF version

Theorem pssdifcom2 4451
Description: Two ways to express non-covering pairs of subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)
Assertion
Ref Expression
pssdifcom2 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊊ (𝐶𝐴) ↔ 𝐴 ⊊ (𝐶𝐵)))

Proof of Theorem pssdifcom2
StepHypRef Expression
1 ssconb 4096 . . . 4 ((𝐵𝐶𝐴𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
21ancoms 463 . . 3 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
3 difcom 4449 . . . . 5 ((𝐶𝐴) ⊆ 𝐵 ↔ (𝐶𝐵) ⊆ 𝐴)
43notbii 323 . . . 4 (¬ (𝐶𝐴) ⊆ 𝐵 ↔ ¬ (𝐶𝐵) ⊆ 𝐴)
54a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → (¬ (𝐶𝐴) ⊆ 𝐵 ↔ ¬ (𝐶𝐵) ⊆ 𝐴))
62, 5anbi12d 643 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐵 ⊆ (𝐶𝐴) ∧ ¬ (𝐶𝐴) ⊆ 𝐵) ↔ (𝐴 ⊆ (𝐶𝐵) ∧ ¬ (𝐶𝐵) ⊆ 𝐴)))
7 dfpss3 4043 . 2 (𝐵 ⊊ (𝐶𝐴) ↔ (𝐵 ⊆ (𝐶𝐴) ∧ ¬ (𝐶𝐴) ⊆ 𝐵))
8 dfpss3 4043 . 2 (𝐴 ⊊ (𝐶𝐵) ↔ (𝐴 ⊆ (𝐶𝐵) ∧ ¬ (𝐶𝐵) ⊆ 𝐴))
96, 7, 83bitr4g 317 1 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊊ (𝐶𝐴) ↔ 𝐴 ⊊ (𝐶𝐵)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wa 400  cdif 3902  wss 3905  wpss 3906
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ne 2959  df-v 3457  df-dif 3908  df-un 3910  df-ss 3922  df-pss 3925
This theorem is referenced by:  fin2i2  10297
  Copyright terms: Public domain W3C validator