Theorem sspsstri 3372

Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004.)

Assertion

Ref	Expression
sspsstri	⊢ ((A ⊆ B ∨ B ⊆ A) ↔ (A ⊊ B ∨ A = B ∨ B ⊊ A))

Proof of Theorem sspsstri

Step	Hyp	Ref	Expression
1		or32 513	. 2 ⊢ (((A ⊊ B ∨ B ⊊ A) ∨ A = B) ↔ ((A ⊊ B ∨ A = B) ∨ B ⊊ A))
2		sspss 3369	. . . 4 ⊢ (A ⊆ B ↔ (A ⊊ B ∨ A = B))
3		sspss 3369	. . . . 5 ⊢ (B ⊆ A ↔ (B ⊊ A ∨ B = A))
4		eqcom 2355	. . . . . 6 ⊢ (B = A ↔ A = B)
5	4	orbi2i 505	. . . . 5 ⊢ ((B ⊊ A ∨ B = A) ↔ (B ⊊ A ∨ A = B))
6	3, 5	bitri 240	. . . 4 ⊢ (B ⊆ A ↔ (B ⊊ A ∨ A = B))
7	2, 6	orbi12i 507	. . 3 ⊢ ((A ⊆ B ∨ B ⊆ A) ↔ ((A ⊊ B ∨ A = B) ∨ (B ⊊ A ∨ A = B)))
8		orordir 517	. . 3 ⊢ (((A ⊊ B ∨ B ⊊ A) ∨ A = B) ↔ ((A ⊊ B ∨ A = B) ∨ (B ⊊ A ∨ A = B)))
9	7, 8	bitr4i 243	. 2 ⊢ ((A ⊆ B ∨ B ⊆ A) ↔ ((A ⊊ B ∨ B ⊊ A) ∨ A = B))
10		df-3or 935	. 2 ⊢ ((A ⊊ B ∨ A = B ∨ B ⊊ A) ↔ ((A ⊊ B ∨ A = B) ∨ B ⊊ A))
11	1, 9, 10	3bitr4i 268	1 ⊢ ((A ⊆ B ∨ B ⊆ A) ↔ (A ⊊ B ∨ A = B ∨ B ⊊ A))

Syntax hints:   ↔ wb 176   ∨ wo 357   ∨ w3o 933   = wceq 1642   ⊆ wss 3258   ⊊ wpss 3259

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3or 935  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260  df-pss 3262

This theorem is referenced by: (None)