Theorem gomaex3lem1 914

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

Hypothesis

Ref	Expression
gomaex3lem1.3	c ≤ d⊥

Assertion

Ref	Expression
gomaex3lem1	(c ∪ (c ∪ d)⊥ ) = d⊥

Proof of Theorem gomaex3lem1

Step	Hyp	Ref	Expression
1		comid 187	. . . 4 c C c
2	1	comcom2 183	. . 3 c C c⊥
3		gomaex3lem1.3	. . . 4 c ≤ d⊥
4	3	lecom 180	. . 3 c C d⊥
5	2, 4	fh3 471	. 2 (c ∪ (c⊥ ∩ d⊥ )) = ((c ∪ c⊥ ) ∩ (c ∪ d⊥ ))
6		anor3 90	. . 3 (c⊥ ∩ d⊥ ) = (c ∪ d)⊥
7	6	lor 70	. 2 (c ∪ (c⊥ ∩ d⊥ )) = (c ∪ (c ∪ d)⊥ )
8		ancom 74	. . 3 ((c ∪ c⊥ ) ∩ (c ∪ d⊥ )) = ((c ∪ d⊥ ) ∩ (c ∪ c⊥ ))
9	3	df-le2 131	. . . . . 6 (c ∪ d⊥ ) = d⊥
10	9	ax-r1 35	. . . . 5 d⊥ = (c ∪ d⊥ )
11		df-t 41	. . . . 5 1 = (c ∪ c⊥ )
12	10, 11	2an 79	. . . 4 (d⊥ ∩ 1) = ((c ∪ d⊥ ) ∩ (c ∪ c⊥ ))
13	12	ax-r1 35	. . 3 ((c ∪ d⊥ ) ∩ (c ∪ c⊥ )) = (d⊥ ∩ 1)
14		an1 106	. . 3 (d⊥ ∩ 1) = d⊥
15	8, 13, 14	3tr 65	. 2 ((c ∪ c⊥ ) ∩ (c ∪ d⊥ )) = d⊥
16	5, 7, 15	3tr2 64	1 (c ∪ (c ∪ d)⊥ ) = d⊥

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