Theorem comdr 466

Description: Commutation dual. Kalmbach 83 p. 23. (Contributed by NM, 27-Aug-1997.)

Hypothesis

Ref	Expression
comdr.1	a = ((a ∪ b) ∩ (a ∪ b⊥ ))

Assertion

Ref	Expression
comdr	a C b

Proof of Theorem comdr

Step	Hyp	Ref	Expression
1		comdr.1	. . . . 5 a = ((a ∪ b) ∩ (a ∪ b⊥ ))
2		df-a 40	. . . . . 6 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )⊥
3		oran 87	. . . . . . . . 9 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
4	3	con2 67	. . . . . . . 8 (a ∪ b)⊥ = (a⊥ ∩ b⊥ )
5		oran 87	. . . . . . . . 9 (a ∪ b⊥ ) = (a⊥ ∩ b⊥ ⊥ )⊥
6	5	con2 67	. . . . . . . 8 (a ∪ b⊥ )⊥ = (a⊥ ∩ b⊥ ⊥ )
7	4, 6	2or 72	. . . . . . 7 ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ ) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))
8	7	ax-r4 37	. . . . . 6 ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )⊥ = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥
9	2, 8	ax-r2 36	. . . . 5 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥
10	1, 9	ax-r2 36	. . . 4 a = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥
11	10	con2 67	. . 3 a⊥ = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))
12	11	df-c1 132	. 2 a⊥ C b⊥
13	12	comcom5 458	1 a C b

This proof depends on syntax axioms:   = wb 1   C wc 3  ^⊥ wn 4   ∪ wo 6   ∩ wa 7

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  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: (None)