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Theorem pssdifcom2 4495
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 4137 . . . 4 ((𝐵𝐶𝐴𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
21ancoms 457 . . 3 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
3 difcom 4493 . . . . 5 ((𝐶𝐴) ⊆ 𝐵 ↔ (𝐶𝐵) ⊆ 𝐴)
43notbii 319 . . . 4 (¬ (𝐶𝐴) ⊆ 𝐵 ↔ ¬ (𝐶𝐵) ⊆ 𝐴)
54a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → (¬ (𝐶𝐴) ⊆ 𝐵 ↔ ¬ (𝐶𝐵) ⊆ 𝐴))
62, 5anbi12d 630 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐵 ⊆ (𝐶𝐴) ∧ ¬ (𝐶𝐴) ⊆ 𝐵) ↔ (𝐴 ⊆ (𝐶𝐵) ∧ ¬ (𝐶𝐵) ⊆ 𝐴)))
7 dfpss3 4085 . 2 (𝐵 ⊊ (𝐶𝐴) ↔ (𝐵 ⊆ (𝐶𝐴) ∧ ¬ (𝐶𝐴) ⊆ 𝐵))
8 dfpss3 4085 . 2 (𝐴 ⊊ (𝐶𝐵) ↔ (𝐴 ⊆ (𝐶𝐵) ∧ ¬ (𝐶𝐵) ⊆ 𝐴))
96, 7, 83bitr4g 313 1 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊊ (𝐶𝐴) ↔ 𝐴 ⊊ (𝐶𝐵)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 394  cdif 3944  wss 3947  wpss 3948
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2697
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2704  df-cleq 2718  df-clel 2803  df-ne 2931  df-v 3464  df-dif 3950  df-un 3952  df-ss 3964  df-pss 3967
This theorem is referenced by:  fin2i2  10361
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