Theorem kb10iii 893

Description: Exercise 10(iii) of Kalmbach p. 30 (in a rewritten form). (Contributed by NM, 9-Jan-2004.)

Hypothesis

Ref	Expression
kb10iii.1	b⊥ ≤ (a →1 c)

Assertion

Ref	Expression
kb10iii	c⊥ ≤ (a →1 b)

Proof of Theorem kb10iii

Step	Hyp	Ref	Expression
1		ud1lem0c 277	. . 3 (a →1 b)⊥ = (a ∩ (a⊥ ∪ b⊥ ))
2		omln 446	. . . . . . . 8 (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ ))) = (a⊥ ∪ b⊥ )
3		u1lem9b 778	. . . . . . . . 9 a⊥ ≤ (a →1 c)
4		kb10iii.1	. . . . . . . . 9 b⊥ ≤ (a →1 c)
5	3, 4	lel2or 170	. . . . . . . 8 (a⊥ ∪ b⊥ ) ≤ (a →1 c)
6	2, 5	bltr 138	. . . . . . 7 (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ ))) ≤ (a →1 c)
7	6	lelan 167	. . . . . 6 (a ∩ (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ )))) ≤ (a ∩ (a →1 c))
8		ancom 74	. . . . . 6 (a ∩ (a →1 c)) = ((a →1 c) ∩ a)
9	7, 8	lbtr 139	. . . . 5 (a ∩ (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ )))) ≤ ((a →1 c) ∩ a)
10		womaon 221	. . . . 5 (a ∩ (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ )))) = (a ∩ (a⊥ ∪ b⊥ ))
11		u1lemaa 600	. . . . 5 ((a →1 c) ∩ a) = (a ∩ c)
12	9, 10, 11	le3tr2 141	. . . 4 (a ∩ (a⊥ ∪ b⊥ )) ≤ (a ∩ c)
13		lear 161	. . . 4 (a ∩ c) ≤ c
14	12, 13	letr 137	. . 3 (a ∩ (a⊥ ∪ b⊥ )) ≤ c
15	1, 14	bltr 138	. 2 (a →1 b)⊥ ≤ c
16	15	lecon2 156	1 c⊥ ≤ (a →1 b)

This proof depends on syntax axioms:   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)