Theorem pssdifcom1 4006

Description: Two ways to express overlapping subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)

Assertion

Ref	Expression
pssdifcom1	⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊊ 𝐴))

Proof of Theorem pssdifcom1

Step	Hyp	Ref	Expression
1		difcom 4005	. . . 4 ⊢ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴)
2	1	a1i 11	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴))
3		ssconb 3705	. . . . 5 ⊢ ((𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶) → (𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
4	3	ancoms 468	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
5	4	notbid 307	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (¬ 𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
6	2, 5	anbi12d 743	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)) ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵))))
7		dfpss3 3655	. 2 ⊢ ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)))
8		dfpss3 3655	. 2 ⊢ ((𝐶 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
9	6, 7, 8	3bitr4g 302	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊊ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∧ wa 383   ∖ cdif 3537   ⊆ wss 3540   ⊊ wpss 3541

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-pss 3556

This theorem is referenced by:  isfin2-2  9024  compssiso  9079