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Theorem pssdifcom1 4418
Description: Two ways to express overlapping subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)
Assertion
Ref Expression
pssdifcom1 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐴) ⊊ 𝐵 ↔ (𝐶𝐵) ⊊ 𝐴))

Proof of Theorem pssdifcom1
StepHypRef Expression
1 difcom 4417 . . . 4 ((𝐶𝐴) ⊆ 𝐵 ↔ (𝐶𝐵) ⊆ 𝐴)
21a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐴) ⊆ 𝐵 ↔ (𝐶𝐵) ⊆ 𝐴))
3 ssconb 4100 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
43ancoms 462 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
54notbid 321 . . 3 ((𝐴𝐶𝐵𝐶) → (¬ 𝐵 ⊆ (𝐶𝐴) ↔ ¬ 𝐴 ⊆ (𝐶𝐵)))
62, 5anbi12d 633 . 2 ((𝐴𝐶𝐵𝐶) → (((𝐶𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶𝐴)) ↔ ((𝐶𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶𝐵))))
7 dfpss3 4049 . 2 ((𝐶𝐴) ⊊ 𝐵 ↔ ((𝐶𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶𝐴)))
8 dfpss3 4049 . 2 ((𝐶𝐵) ⊊ 𝐴 ↔ ((𝐶𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶𝐵)))
96, 7, 83bitr4g 317 1 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐴) ⊊ 𝐵 ↔ (𝐶𝐵) ⊊ 𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wa 399  cdif 3916  wss 3919  wpss 3920
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-ne 3015  df-v 3482  df-dif 3922  df-un 3924  df-in 3926  df-ss 3936  df-pss 3938
This theorem is referenced by:  isfin2-2  9739  compssiso  9794
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