Theorem gomaex3lem3 896

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

Assertion

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

Proof of Theorem gomaex3lem3

Step	Hyp	Ref	Expression
1		anor1 80	. . . . 5 (p⊥ ∩ (p⊥ ∩ q)⊥ ) = (p⊥ ⊥ ∪ (p⊥ ∩ q))⊥
2	1	ax-r1 34	. . . 4 (p⊥ ⊥ ∪ (p⊥ ∩ q))⊥ = (p⊥ ∩ (p⊥ ∩ q)⊥ )
3		df-i1 43	. . . . 5 (p⊥ →1 q) = (p⊥ ⊥ ∪ (p⊥ ∩ q))
4	3	ax-r4 36	. . . 4 (p⊥ →1 q)⊥ = (p⊥ ⊥ ∪ (p⊥ ∩ q))⊥
5		id 58	. . . 4 (p⊥ ∩ (p⊥ ∩ q)⊥ ) = (p⊥ ∩ (p⊥ ∩ q)⊥ )
6	2, 4, 5	3tr1 60	. . 3 (p⊥ →1 q)⊥ = (p⊥ ∩ (p⊥ ∩ q)⊥ )
7	6	ax-r5 37	. 2 ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q)) = ((p⊥ ∩ (p⊥ ∩ q)⊥ ) ∪ (p⊥ ∩ q))
8		coman1 177	. . 3 (p⊥ ∩ q) C p⊥
9		comid 179	. . . 4 (p⊥ ∩ q) C (p⊥ ∩ q)
10	9	comcom2 175	. . 3 (p⊥ ∩ q) C (p⊥ ∩ q)⊥
11	8, 10	fh3r 457	. 2 ((p⊥ ∩ (p⊥ ∩ q)⊥ ) ∪ (p⊥ ∩ q)) = ((p⊥ ∪ (p⊥ ∩ q)) ∩ ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q)))
12		a5b 112	. . . 4 (p⊥ ∪ (p⊥ ∩ q)) = p⊥
13		ax-a2 30	. . . . 5 ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q)) = ((p⊥ ∩ q) ∪ (p⊥ ∩ q)⊥ )
14		df-t 40	. . . . . 6 1 = ((p⊥ ∩ q) ∪ (p⊥ ∩ q)⊥ )
15	14	ax-r1 34	. . . . 5 ((p⊥ ∩ q) ∪ (p⊥ ∩ q)⊥ ) = 1
16	13, 15	ax-r2 35	. . . 4 ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q)) = 1
17	12, 16	2an 72	. . 3 ((p⊥ ∪ (p⊥ ∩ q)) ∩ ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q))) = (p⊥ ∩ 1)
18		an1 98	. . 3 (p⊥ ∩ 1) = p⊥
19	17, 18	ax-r2 35	. 2 ((p⊥ ∪ (p⊥ ∩ q)) ∩ ((p⊥ ∩ q)⊥ ∪ (p⊥ ∩ q))) = p⊥
20	7, 11, 19	3tr 62	1 ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q)) = p⊥

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₁ wi1 13

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-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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