Theorem sspsstri 3671

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 548	. 2 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ 𝐵 ⊊ 𝐴))
2		sspss 3668	. . . 4 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵))
3		sspss 3668	. . . . 5 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ⊊ 𝐴 ∨ 𝐵 = 𝐴))
4		eqcom 2617	. . . . . 6 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	orbi2i 540	. . . . 5 ⊢ ((𝐵 ⊊ 𝐴 ∨ 𝐵 = 𝐴) ↔ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵))
6	3, 5	bitri 263	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵))
7	2, 6	orbi12i 542	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵)))
8		orordir 552	. . 3 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵)))
9	7, 8	bitr4i 266	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵))
10		df-3or 1032	. 2 ⊢ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ 𝐵 ⊊ 𝐴))
11	1, 9, 10	3bitr4i 291	1 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴))

Syntax hints:   ↔ wb 195   ∨ wo 382   ∨ w3o 1030   = wceq 1475   ⊆ 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-3or 1032  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-ne 2782  df-in 3547  df-ss 3554  df-pss 3556

This theorem is referenced by:  ordtri3or  5672  sorpss  6840  sorpssi  6841  funpsstri  30909