Theorem gomaex3lem2 895

Description: Lemma for Godowski 6-var -> Mayet Example 3.

Hypothesis

Ref	Expression
gomaex3lem2.5	e ≤ f⊥

Assertion

Ref	Expression
gomaex3lem2	((e ∪ f)⊥ ∪ f) = e⊥

Proof of Theorem gomaex3lem2

Step	Hyp	Ref	Expression
1		gomaex3lem2.5	. . . . . 6 e ≤ f⊥
2	1	lecon3 149	. . . . 5 f ≤ e⊥
3	2	lecom 172	. . . 4 f C e⊥
4		comid 179	. . . . 5 f C f
5	4	comcom2 175	. . . 4 f C f⊥
6	3, 5	fh3r 457	. . 3 ((e⊥ ∩ f⊥ ) ∪ f) = ((e⊥ ∪ f) ∩ (f⊥ ∪ f))
7		anor3 82	. . . . 5 (e⊥ ∩ f⊥ ) = (e ∪ f)⊥
8	7	ax-r5 37	. . . 4 ((e⊥ ∩ f⊥ ) ∪ f) = ((e ∪ f)⊥ ∪ f)
9	8	ax-r1 34	. . 3 ((e ∪ f)⊥ ∪ f) = ((e⊥ ∩ f⊥ ) ∪ f)
10		a5c 113	. . . . . 6 (e⊥ ∩ (e⊥ ∪ f)) = e⊥
11	10	df2le1 127	. . . . 5 e⊥ ≤ (e⊥ ∪ f)
12		leid 140	. . . . . 6 e⊥ ≤ e⊥
13	12, 2	lel2or 162	. . . . 5 (e⊥ ∪ f) ≤ e⊥
14	11, 13	lebi 137	. . . 4 e⊥ = (e⊥ ∪ f)
15		df-t 40	. . . . 5 1 = (f ∪ f⊥ )
16		ax-a2 30	. . . . 5 (f ∪ f⊥ ) = (f⊥ ∪ f)
17	15, 16	ax-r2 35	. . . 4 1 = (f⊥ ∪ f)
18	14, 17	2an 72	. . 3 (e⊥ ∩ 1) = ((e⊥ ∪ f) ∩ (f⊥ ∪ f))
19	6, 9, 18	3tr1 60	. 2 ((e ∪ f)⊥ ∪ f) = (e⊥ ∩ 1)
20		an1 98	. 2 (e⊥ ∩ 1) = e⊥
21	19, 20	ax-r2 35	1 ((e ∪ f)⊥ ∪ f) = e⊥

Syntax hints:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9

This theorem is referenced by:  gomaex3lem7 900

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-le1 122  df-le2 123  df-c1 124  df-c2 125

metamath.org