Theorem pssdifcom1 4489

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 4488	. . . 4 ⊢ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴)
2	1	a1i 11	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴))
3		ssconb 4137	. . . . 5 ⊢ ((𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶) → (𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
4	3	ancoms 458	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
5	4	notbid 318	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (¬ 𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
6	2, 5	anbi12d 630	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)) ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵))))
7		dfpss3 4086	. 2 ⊢ ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)))
8		dfpss3 4086	. 2 ⊢ ((𝐶 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
9	6, 7, 8	3bitr4g 314	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊊ 𝐴))

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