Theorem nomcon0 301

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

Assertion

Ref	Expression
nomcon0	(a ≡0 b) = (b⊥ ≡0 a⊥ )

Proof of Theorem nomcon0

Step	Hyp	Ref	Expression
1		ax-a2 31	. . . 4 (a⊥ ∪ b) = (b ∪ a⊥ )
2		ax-a1 30	. . . . 5 b = b⊥ ⊥
3	2	ax-r5 38	. . . 4 (b ∪ a⊥ ) = (b⊥ ⊥ ∪ a⊥ )
4	1, 3	ax-r2 36	. . 3 (a⊥ ∪ b) = (b⊥ ⊥ ∪ a⊥ )
5		ax-a2 31	. . . 4 (b⊥ ∪ a) = (a ∪ b⊥ )
6		ax-a1 30	. . . . 5 a = a⊥ ⊥
7	6	ax-r5 38	. . . 4 (a ∪ b⊥ ) = (a⊥ ⊥ ∪ b⊥ )
8	5, 7	ax-r2 36	. . 3 (b⊥ ∪ a) = (a⊥ ⊥ ∪ b⊥ )
9	4, 8	2an 79	. 2 ((a⊥ ∪ b) ∩ (b⊥ ∪ a)) = ((b⊥ ⊥ ∪ a⊥ ) ∩ (a⊥ ⊥ ∪ b⊥ ))
10		df-id0 49	. 2 (a ≡0 b) = ((a⊥ ∪ b) ∩ (b⊥ ∪ a))
11		df-id0 49	. 2 (b⊥ ≡0 a⊥ ) = ((b⊥ ⊥ ∪ a⊥ ) ∩ (a⊥ ⊥ ∪ b⊥ ))
12	9, 10, 11	3tr1 63	1 (a ≡0 b) = (b⊥ ≡0 a⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   ≡₀ wid0 17

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-id0 49

This theorem is used by:  nom50  331