Theorem omlem1 127

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

Assertion

Ref	Expression
omlem1	((a ∪ (a⊥ ∩ (a ∪ b))) ∪ (a ∪ b)) = (a ∪ b)

Proof of Theorem omlem1

Step	Hyp	Ref	Expression
1		ax-a2 31	. . 3 ((a ∪ (a⊥ ∩ (a ∪ b))) ∪ (a ∪ b)) = ((a ∪ b) ∪ (a ∪ (a⊥ ∩ (a ∪ b))))
2		ax-a3 32	. . 3 (((a ∪ (a⊥ ∩ (a ∪ b))) ∪ a) ∪ b) = ((a ∪ (a⊥ ∩ (a ∪ b))) ∪ (a ∪ b))
3		ax-a3 32	. . 3 (((a ∪ b) ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = ((a ∪ b) ∪ (a ∪ (a⊥ ∩ (a ∪ b))))
4	1, 2, 3	3tr1 63	. 2 (((a ∪ (a⊥ ∩ (a ∪ b))) ∪ a) ∪ b) = (((a ∪ b) ∪ a) ∪ (a⊥ ∩ (a ∪ b)))
5		ax-a3 32	. . . . . . 7 ((a ∪ a) ∪ b) = (a ∪ (a ∪ b))
6		ax-a2 31	. . . . . . 7 (a ∪ (a ∪ b)) = ((a ∪ b) ∪ a)
7	5, 6	ax-r2 36	. . . . . 6 ((a ∪ a) ∪ b) = ((a ∪ b) ∪ a)
8	7	ax-r1 35	. . . . 5 ((a ∪ b) ∪ a) = ((a ∪ a) ∪ b)
9		oridm 110	. . . . . 6 (a ∪ a) = a
10	9	ax-r5 38	. . . . 5 ((a ∪ a) ∪ b) = (a ∪ b)
11	8, 10	ax-r2 36	. . . 4 ((a ∪ b) ∪ a) = (a ∪ b)
12		ancom 74	. . . 4 (a⊥ ∩ (a ∪ b)) = ((a ∪ b) ∩ a⊥ )
13	11, 12	2or 72	. . 3 (((a ∪ b) ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = ((a ∪ b) ∪ ((a ∪ b) ∩ a⊥ ))
14		orabs 120	. . 3 ((a ∪ b) ∪ ((a ∪ b) ∩ a⊥ )) = (a ∪ b)
15	13, 14	ax-r2 36	. 2 (((a ∪ b) ∪ a) ∪ (a⊥ ∩ (a ∪ b))) = (a ∪ b)
16	4, 2, 15	3tr2 64	1 ((a ∪ (a⊥ ∩ (a ∪ b))) ∪ (a ∪ b)) = (a ∪ b)

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

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42

This theorem is referenced by:  woml  211  oml  445