Theorem pssdifcom1 4499

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 4498	. . . 4 ⊢ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴)
2	1	a1i 11	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴))
3		ssconb 4155	. . . . 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 632	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)) ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵))))
7		dfpss3 4102	. 2 ⊢ ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)))
8		dfpss3 4102	. 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 206   ∧ wa 395   ∖ cdif 3963   ⊆ wss 3966   ⊊ wpss 3967

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1542  df-ex 1779  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ne 2941  df-v 3483  df-dif 3969  df-un 3971  df-ss 3983  df-pss 3986

This theorem is referenced by:  isfin2-2  10366  compssiso  10421