Theorem pssdifcom1 4449

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 4448	. . . 4 ⊢ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴)
2	1	a1i 11	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊆ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊆ 𝐴))
3		ssconb 4095	. . . . 5 ⊢ ((𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶) → (𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
4	3	ancoms 463	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
5	4	notbid 321	. . 3 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (¬ 𝐵 ⊆ (𝐶 ∖ 𝐴) ↔ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
6	2, 5	anbi12d 643	. 2 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)) ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵))))
7		dfpss3 4042	. 2 ⊢ ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ ((𝐶 ∖ 𝐴) ⊆ 𝐵 ∧ ¬ 𝐵 ⊆ (𝐶 ∖ 𝐴)))
8		dfpss3 4042	. 2 ⊢ ((𝐶 ∖ 𝐵) ⊊ 𝐴 ↔ ((𝐶 ∖ 𝐵) ⊆ 𝐴 ∧ ¬ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
9	6, 7, 8	3bitr4g 317	1 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊊ 𝐴))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   ∖ cdif 3901   ⊆ wss 3904   ⊊ wpss 3905

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-v 3456  df-dif 3907  df-un 3909  df-ss 3921  df-pss 3924

This theorem is used by:  isfin2-2  10309  compssiso  10364