Theorem sspsstri 2138

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 2135	. . 3 |- (A (_ B <-> (A (. B \/ A = B))
2		sspss 2135	. . . 4 |- (B (_ A <-> (B (. A \/ B = A))
3		eqcom 1469	. . . . 5 |- (B = A <-> A = B)
4	3	orbi2i 255	. . . 4 |- ((B (. A \/ B = A) <-> (B (. A \/ A = B))
5	2, 4	bitr 173	. . 3 |- (B (_ A <-> (B (. A \/ A = B))
6	1, 5	orbi12i 257	. 2 |- ((A (_ B \/ B (_ A) <-> ((A (. B \/ A = B) \/ (B (. A \/ A = B)))
7		orordir 267	. 2 |- (((A (. B \/ B (. A) \/ A = B) <-> ((A (. B \/ A = B) \/ (B (. A \/ A = B)))
8		or23 263	. . 3 |- (((A (. B \/ B (. A) \/ A = B) <-> ((A (. B \/ A = B) \/ B (. A))
9		df-3or 774	. . 3 |- ((A (. B \/ A = B \/ B (. A) <-> ((A (. B \/ A = B) \/ B (. A))
10	8, 9	bitr4 176	. 2 |- (((A (. B \/ B (. A) \/ A = B) <-> (A (. B \/ A = B \/ B (. A))
11	6, 7, 10	3bitr2 179	1 |- ((A (_ B \/ B (_ A) <-> (A (. B \/ A = B \/ B (. A))

Syntax hints:   <-> wb 146   \/ wo 222   \/ w3o 772   = wceq 953   (_ wss 2037   (. wpss 2038

This theorem is referenced by:  zorn 4769

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-10 963  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-16 1206  ax-11o 1213  ax-ext 1452

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3or 774  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-ne 1579  df-in 2041  df-ss 2043  df-pss 2045

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