Theorem i3lem3 488

Description: Lemma for Kalmbach implication.

Hypothesis

Ref	Expression
i3lem.1	(a →3 b) = 1

Assertion

Ref	Expression
i3lem3	((a⊥ ∪ b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Proof of Theorem i3lem3

Step	Hyp	Ref	Expression
1		omlan 430	. 2 (b⊥ ∩ (b ∪ (b⊥ ∩ a⊥ ))) = (b⊥ ∩ a⊥ )
2		ancom 68	. . 3 ((a⊥ ∪ b) ∩ b⊥ ) = (b⊥ ∩ (a⊥ ∪ b))
3		ax-a2 30	. . . . 5 (a⊥ ∪ b) = (b ∪ a⊥ )
4		ax-a3 31	. . . . . . 7 ((b ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (b ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
5	4	ax-r1 34	. . . . . 6 (b ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = ((b ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
6		i3lem.1	. . . . . . . 8 (a →3 b) = 1
7	6	i3lem1 486	. . . . . . 7 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = a⊥
8	7	lor 66	. . . . . 6 (b ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (b ∪ a⊥ )
9		ancom 68	. . . . . . . . 9 (a⊥ ∩ b) = (b ∩ a⊥ )
10	9	lor 66	. . . . . . . 8 (b ∪ (a⊥ ∩ b)) = (b ∪ (b ∩ a⊥ ))
11		a5b 112	. . . . . . . 8 (b ∪ (b ∩ a⊥ )) = b
12	10, 11	ax-r2 35	. . . . . . 7 (b ∪ (a⊥ ∩ b)) = b
13		ancom 68	. . . . . . 7 (a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ )
14	12, 13	2or 67	. . . . . 6 ((b ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (b ∪ (b⊥ ∩ a⊥ ))
15	5, 8, 14	3tr2 61	. . . . 5 (b ∪ a⊥ ) = (b ∪ (b⊥ ∩ a⊥ ))
16	3, 15	ax-r2 35	. . . 4 (a⊥ ∪ b) = (b ∪ (b⊥ ∩ a⊥ ))
17	16	lan 70	. . 3 (b⊥ ∩ (a⊥ ∪ b)) = (b⊥ ∩ (b ∪ (b⊥ ∩ a⊥ )))
18	2, 17	ax-r2 35	. 2 ((a⊥ ∪ b) ∩ b⊥ ) = (b⊥ ∩ (b ∪ (b⊥ ∩ a⊥ )))
19	1, 18, 13	3tr1 60	1 ((a⊥ ∪ b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₃ wi3 15

This theorem is referenced by:  i3le 497

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  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  df-i3 45  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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