Theorem sspsstri 1572

Description: Two ways of stating trichotomy with respect to inclusion.

Assertion

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

Proof of Theorem sspsstri

Step	Hyp	Ref	Expression
1		sspss 1569	. . 3 ⊢ (A ⊆ B ↔ (A ⊂ B ∨ A = B))
2		sspss 1569	. . . 4 ⊢ (B ⊆ A ↔ (B ⊂ A ∨ B = A))
3		cleqcom 1103	. . . . 5 ⊢ (B = A ↔ A = B)
4	3	orbi2i 214	. . . 4 ⊢ ((B ⊂ A ∨ B = A) ↔ (B ⊂ A ∨ A = B))
5	2, 4	bitr 151	. . 3 ⊢ (B ⊆ A ↔ (B ⊂ A ∨ A = B))
6	1, 5	orbi12i 216	. 2 ⊢ ((A ⊆ B ∨ B ⊆ A) ↔ ((A ⊂ B ∨ A = B) ∨ (B ⊂ A ∨ A = B)))
7		orordir 223	. . 3 ⊢ (((A ⊂ B ∨ B ⊂ A) ∨ A = B) ↔ ((A ⊂ B ∨ A = B) ∨ (B ⊂ A ∨ A = B)))
8		or23 219	. . . 4 ⊢ (((A ⊂ B ∨ B ⊂ A) ∨ A = B) ↔ ((A ⊂ B ∨ A = B) ∨ B ⊂ A))
9		df-3or 582	. . . 4 ⊢ ((A ⊂ B ∨ A = B ∨ B ⊂ A) ↔ ((A ⊂ B ∨ A = B) ∨ B ⊂ A))
10	8, 9	bitr4 154	. . 3 ⊢ (((A ⊂ B ∨ B ⊂ A) ∨ A = B) ↔ (A ⊂ B ∨ A = B ∨ B ⊂ A))
11	7, 10	bitr3 153	. 2 ⊢ (((A ⊂ B ∨ A = B) ∨ (B ⊂ A ∨ A = B)) ↔ (A ⊂ B ∨ A = B ∨ B ⊂ A))
12	6, 11	bitr 151	1 ⊢ ((A ⊆ B ∨ B ⊆ A) ↔ (A ⊂ B ∨ A = B ∨ B ⊂ A))

Syntax hints:   ↔ wb 127   ∨ wo 195   ∨ w3o 580   = wceq 1091   ⊆ wss 1487   ⊂ wpss 1488

This theorem is referenced by:  zorn2 3612

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ne 1192  df-in 1491  df-ss 1492  df-pss 1494

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