Theorem gomaex3lem3 916

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

Assertion

Ref	Expression
gomaex3lem3	((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q)) = p⊥

Proof of Theorem gomaex3lem3

Step	Hyp	Ref	Expression
1		anor1 88	. . . . 5 (p⊥ ∩ (p⊥ ∩ q)⊥ ) = (p⊥ ⊥ ∪ (p⊥ ∩ q))⊥
2	1	ax-r1 35	. . . 4 (p⊥ ⊥ ∪ (p⊥ ∩ q))⊥ = (p⊥ ∩ (p⊥ ∩ q)⊥ )
3		df-i1 44	. . . . 5 (p⊥ →1 q) = (p⊥ ⊥ ∪ (p⊥ ∩ q))
4	3	ax-r4 37	. . . 4 (p⊥ →1 q)⊥ = (p⊥ ⊥ ∪ (p⊥ ∩ q))⊥
5		id 59	. . . 4 (p⊥ ∩ (p⊥ ∩ q)⊥ ) = (p⊥ ∩ (p⊥ ∩ q)⊥ )
6	2, 4, 5	3tr1 63	. . 3 (p⊥ →1 q)⊥ = (p⊥ ∩ (p⊥ ∩ q)⊥ )
7	6	ax-r5 38	. 2 ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q)) = ((p⊥ ∩ (p⊥ ∩ q)⊥ ) ∪ (p⊥ ∩ q))
8		coman1 185	. . 3 (p⊥ ∩ q) C p⊥
9		comid 187	. . . 4 (p⊥ ∩ q) C (p⊥ ∩ q)
10	9	comcom2 183	. . 3 (p⊥ ∩ q) C (p⊥ ∩ q)⊥
11	8, 10	fh3r 475	. 2 ((p⊥ ∩ (p⊥ ∩ q)⊥ ) ∪ (p⊥ ∩ q)) = ((p⊥ ∪ (p⊥ ∩ q)) ∩ ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q)))
12		orabs 120	. . . 4 (p⊥ ∪ (p⊥ ∩ q)) = p⊥
13		ax-a2 31	. . . . 5 ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q)) = ((p⊥ ∩ q) ∪ (p⊥ ∩ q)⊥ )
14		df-t 41	. . . . . 6 1 = ((p⊥ ∩ q) ∪ (p⊥ ∩ q)⊥ )
15	14	ax-r1 35	. . . . 5 ((p⊥ ∩ q) ∪ (p⊥ ∩ q)⊥ ) = 1
16	13, 15	ax-r2 36	. . . 4 ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q)) = 1
17	12, 16	2an 79	. . 3 ((p⊥ ∪ (p⊥ ∩ q)) ∩ ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q))) = (p⊥ ∩ 1)
18		an1 106	. . 3 (p⊥ ∩ 1) = p⊥
19	17, 18	ax-r2 36	. 2 ((p⊥ ∪ (p⊥ ∩ q)) ∩ ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q))) = p⊥
20	7, 11, 19	3tr 65	1 ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q)) = p⊥

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12

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-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  gomaex3lem7  920