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Theorem pssdifcom2 4456
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 4104 . . . 4 ((𝐵𝐶𝐴𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
21ancoms 463 . . 3 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
3 difcom 4454 . . . . 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 4051 . 2 (𝐵 ⊊ (𝐶𝐴) ↔ (𝐵 ⊆ (𝐶𝐴) ∧ ¬ (𝐶𝐴) ⊆ 𝐵))
8 dfpss3 4051 . 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 3910  wss 3913  wpss 3914
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ne 2965  df-v 3465  df-dif 3916  df-un 3918  df-ss 3930  df-pss 3933
This theorem is referenced by:  fin2i2  10301
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