Theorem omlem2 128

Description: Lemma in proof of Thm. 1 of Pavicic 1987. (Contributed by NM, 12-Aug-1997.)

Assertion

Ref	Expression
omlem2	((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b)))) = 1

Proof of Theorem omlem2

Step	Hyp	Ref	Expression
1		ax-a2 31	. . 3 ((a ∪ b)⊥ ∪ a) = (a ∪ (a ∪ b)⊥ )
2		anor2 89	. . 3 (a⊥ ∩ (a ∪ b)) = (a ∪ (a ∪ b)⊥ )⊥
3	1, 2	2or 72	. 2 (((a ∪ b)⊥ ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = ((a ∪ (a ∪ b)⊥ ) ∪ (a ∪ (a ∪ b)⊥ )⊥ )
4		ax-a3 32	. . 3 (((a ∪ b)⊥ ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = ((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b))))
5	4	ax-r1 35	. 2 ((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b)))) = (((a ∪ b)⊥ ∪ a) ∪ (a⊥ ∩ (a ∪ b)))
6		df-t 41	. 2 1 = ((a ∪ (a ∪ b)⊥ ) ∪ (a ∪ (a ∪ b)⊥ )⊥ )
7	3, 5, 6	3tr1 63	1 ((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b)))) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-t 41

This theorem is used by:  woml  211  wql2lem3  290  oml  445