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Theorem pssdifcom1 4489
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 4488 . . . 4 ((𝐶𝐴) ⊆ 𝐵 ↔ (𝐶𝐵) ⊆ 𝐴)
21a1i 11 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐴) ⊆ 𝐵 ↔ (𝐶𝐵) ⊆ 𝐴))
3 ssconb 4137 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
43ancoms 458 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵 ⊆ (𝐶𝐴) ↔ 𝐴 ⊆ (𝐶𝐵)))
54notbid 318 . . 3 ((𝐴𝐶𝐵𝐶) → (¬ 𝐵 ⊆ (𝐶𝐴) ↔ ¬ 𝐴 ⊆ (𝐶𝐵)))
62, 5anbi12d 630 . 2 ((𝐴𝐶𝐵𝐶) → (((𝐶𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶𝐴)) ↔ ((𝐶𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶𝐵))))
7 dfpss3 4086 . 2 ((𝐶𝐴) ⊊ 𝐵 ↔ ((𝐶𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶𝐴)))
8 dfpss3 4086 . 2 ((𝐶𝐵) ⊊ 𝐴 ↔ ((𝐶𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶𝐵)))
96, 7, 83bitr4g 314 1 ((𝐴𝐶𝐵𝐶) → ((𝐶𝐴) ⊊ 𝐵 ↔ (𝐶𝐵) ⊊ 𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 395  cdif 3945  wss 3948  wpss 3949
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-ne 2940  df-v 3475  df-dif 3951  df-un 3953  df-in 3955  df-ss 3965  df-pss 3967
This theorem is referenced by:  isfin2-2  10320  compssiso  10375
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