Theorem nomcon1 302

Description: Lemma for "Non-Orthomodular Models..." paper. (Contributed by NM, 7-Feb-1999.)

Assertion

Ref	Expression
nomcon1	(a ≡1 b) = (b⊥ ≡2 a⊥ )

Proof of Theorem nomcon1

Step	Hyp	Ref	Expression
1		ax-a2 31	. . . 4 (a ∪ b⊥ ) = (b⊥ ∪ a)
2		ax-a1 30	. . . . 5 a = a⊥ ⊥
3	2	lor 70	. . . 4 (b⊥ ∪ a) = (b⊥ ∪ a⊥ ⊥ )
4	1, 3	ax-r2 36	. . 3 (a ∪ b⊥ ) = (b⊥ ∪ a⊥ ⊥ )
5		ancom 74	. . . . 5 (a ∩ b) = (b ∩ a)
6		ax-a1 30	. . . . . 6 b = b⊥ ⊥
7	6, 2	2an 79	. . . . 5 (b ∩ a) = (b⊥ ⊥ ∩ a⊥ ⊥ )
8	5, 7	ax-r2 36	. . . 4 (a ∩ b) = (b⊥ ⊥ ∩ a⊥ ⊥ )
9	8	lor 70	. . 3 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
10	4, 9	2an 79	. 2 ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b))) = ((b⊥ ∪ a⊥ ⊥ ) ∩ (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )))
11		df-id1 50	. 2 (a ≡1 b) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b)))
12		df-id2 51	. 2 (b⊥ ≡2 a⊥ ) = ((b⊥ ∪ a⊥ ⊥ ) ∩ (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ )))
13	10, 11, 12	3tr1 63	1 (a ≡1 b) = (b⊥ ≡2 a⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   ≡₁ wid1 18   ≡₂ wid2 19

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-id1 50  df-id2 51

This theorem is referenced by:  nomcon4  305  nom51  332