Theorem omlem2 120

Description: Lemma in proof of Th. 1 of Pavicic 1987.

Assertion

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

Proof of Theorem omlem2

Step	Hyp	Ref	Expression
1		ax-a2 30	. . 3 ((a ∪ b)⊥ ∪ a) = (a ∪ (a ∪ b)⊥ )
2		anor2 81	. . 3 (a⊥ ∩ (a ∪ b)) = (a ∪ (a ∪ b)⊥ )⊥
3	1, 2	2or 67	. 2 (((a ∪ b)⊥ ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = ((a ∪ (a ∪ b)⊥ ) ∪ (a ∪ (a ∪ b)⊥ )⊥ )
4		ax-a3 31	. . 3 (((a ∪ b)⊥ ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = ((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b))))
5	4	ax-r1 34	. 2 ((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b)))) = (((a ∪ b)⊥ ∪ a) ∪ (a⊥ ∩ (a ∪ b)))
6		df-t 40	. 2 1 = ((a ∪ (a ∪ b)⊥ ) ∪ (a ∪ (a ∪ b)⊥ )⊥ )
7	3, 5, 6	3tr1 60	1 ((a ∪ b)⊥ ∪ (a ∪ (a⊥ ∩ (a ∪ b)))) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9

This theorem is referenced by:  woml 203  wql2lem3 282  oml 427

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-t 40

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