Theorem omla 429

Description: Orthomodular law.

Assertion

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

Proof of Theorem omla

Step	Hyp	Ref	Expression
1		df-a 39	. . . . . . 7 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥
2		df-a 39	. . . . . . . . . 10 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
3	2	ax-r1 34	. . . . . . . . 9 (a⊥ ∪ b⊥ )⊥ = (a ∩ b)
4	3	lor 66	. . . . . . . 8 (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ ) = (a⊥ ∪ (a ∩ b))
5	4	ax-r4 36	. . . . . . 7 (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥ = (a⊥ ∪ (a ∩ b))⊥
6	1, 5	ax-r2 35	. . . . . 6 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a ∩ b))⊥
7	6	ax-r1 34	. . . . 5 (a⊥ ∪ (a ∩ b))⊥ = (a ∩ (a⊥ ∪ b⊥ ))
8	7	lor 66	. . . 4 (a⊥ ∪ (a⊥ ∪ (a ∩ b))⊥ ) = (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ )))
9		omln 428	. . . 4 (a⊥ ∪ (a ∩ (a⊥ ∪ b⊥ ))) = (a⊥ ∪ b⊥ )
10	8, 9	ax-r2 35	. . 3 (a⊥ ∪ (a⊥ ∪ (a ∩ b))⊥ ) = (a⊥ ∪ b⊥ )
11		df-a 39	. . . 4 (a ∩ (a⊥ ∪ (a ∩ b))) = (a⊥ ∪ (a⊥ ∪ (a ∩ b))⊥ )⊥
12	11	con2 64	. . 3 (a ∩ (a⊥ ∪ (a ∩ b)))⊥ = (a⊥ ∪ (a⊥ ∪ (a ∩ b))⊥ )
13	2	con2 64	. . 3 (a ∩ b)⊥ = (a⊥ ∪ b⊥ )
14	10, 12, 13	3tr1 60	. 2 (a ∩ (a⊥ ∪ (a ∩ b)))⊥ = (a ∩ b)⊥
15	14	con1 63	1 (a ∩ (a⊥ ∪ (a ∩ b))) = (a ∩ b)

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

This theorem is referenced by:  omlan 430  oml5a 432  gsth2 472  oa3-2to2s 970

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41

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