Theorem gomaex3lem2 915

Description: Lemma for Godowski 6-var -> Mayet Example 3. (Contributed by NM, 29-Nov-1999.)

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 157	. . . . 5 f ≤ e⊥
3	2	lecom 180	. . . 4 f C e⊥
4		comid 187	. . . . 5 f C f
5	4	comcom2 183	. . . 4 f C f⊥
6	3, 5	fh3r 475	. . 3 ((e⊥ ∩ f⊥ ) ∪ f) = ((e⊥ ∪ f) ∩ (f⊥ ∪ f))
7		anor3 90	. . . . 5 (e⊥ ∩ f⊥ ) = (e ∪ f)⊥
8	7	ax-r5 38	. . . 4 ((e⊥ ∩ f⊥ ) ∪ f) = ((e ∪ f)⊥ ∪ f)
9	8	ax-r1 35	. . 3 ((e ∪ f)⊥ ∪ f) = ((e⊥ ∩ f⊥ ) ∪ f)
10		anabs 121	. . . . . 6 (e⊥ ∩ (e⊥ ∪ f)) = e⊥
11	10	df2le1 135	. . . . 5 e⊥ ≤ (e⊥ ∪ f)
12		leid 148	. . . . . 6 e⊥ ≤ e⊥
13	12, 2	lel2or 170	. . . . 5 (e⊥ ∪ f) ≤ e⊥
14	11, 13	lebi 145	. . . 4 e⊥ = (e⊥ ∪ f)
15		df-t 41	. . . . 5 1 = (f ∪ f⊥ )
16		ax-a2 31	. . . . 5 (f ∪ f⊥ ) = (f⊥ ∪ f)
17	15, 16	ax-r2 36	. . . 4 1 = (f⊥ ∪ f)
18	14, 17	2an 79	. . 3 (e⊥ ∩ 1) = ((e⊥ ∪ f) ∩ (f⊥ ∪ f))
19	6, 9, 18	3tr1 63	. 2 ((e ∪ f)⊥ ∪ f) = (e⊥ ∩ 1)
20		an1 106	. 2 (e⊥ ∩ 1) = e⊥
21	19, 20	ax-r2 36	1 ((e ∪ f)⊥ ∪ f) = e⊥

This proof depends on syntax axioms:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  gomaex3lem7  920