Theorem wcomd 400

Description: Commutation dual. Kalmbach 83 p. 23.

Hypothesis

Ref	Expression
wcomcom.1	C (a, b) = 1

Assertion

Ref	Expression
wcomd	(a ≡ ((a ∪ b) ∩ (a ∪ b⊥ ))) = 1

Proof of Theorem wcomd

Step	Hyp	Ref	Expression
1		wcomcom.1	. . . . 5 C (a, b) = 1
2	1	wcomcom4 399	. . . 4 C (a⊥ , b⊥ ) = 1
3	2	wdf-c2 366	. . 3 (a⊥ ≡ ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))) = 1
4	3	wcon3 201	. 2 (a ≡ ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥ ) = 1
5		oran 79	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ )) = ((a⊥ ∩ b⊥ )⊥ ∩ (a⊥ ∩ b⊥ ⊥ )⊥ )⊥
6	5	bi1 110	. . . 4 (((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ )) ≡ ((a⊥ ∩ b⊥ )⊥ ∩ (a⊥ ∩ b⊥ ⊥ )⊥ )⊥ ) = 1
7	6	wcon2 200	. . 3 (((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥ ≡ ((a⊥ ∩ b⊥ )⊥ ∩ (a⊥ ∩ b⊥ ⊥ )⊥ )) = 1
8		oran 79	. . . . . 6 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
9	8	bi1 110	. . . . 5 ((a ∪ b) ≡ (a⊥ ∩ b⊥ )⊥ ) = 1
10		oran 79	. . . . . 6 (a ∪ b⊥ ) = (a⊥ ∩ b⊥ ⊥ )⊥
11	10	bi1 110	. . . . 5 ((a ∪ b⊥ ) ≡ (a⊥ ∩ b⊥ ⊥ )⊥ ) = 1
12	9, 11	w2an 355	. . . 4 (((a ∪ b) ∩ (a ∪ b⊥ )) ≡ ((a⊥ ∩ b⊥ )⊥ ∩ (a⊥ ∩ b⊥ ⊥ )⊥ )) = 1
13	12	wr1 189	. . 3 (((a⊥ ∩ b⊥ )⊥ ∩ (a⊥ ∩ b⊥ ⊥ )⊥ ) ≡ ((a ∪ b) ∩ (a ∪ b⊥ ))) = 1
14	7, 13	wr2 353	. 2 (((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥ ≡ ((a ∪ b) ∩ (a ∪ b⊥ ))) = 1
15	4, 14	wr2 353	1 (a ≡ ((a ∪ b) ∩ (a ∪ b⊥ ))) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 9   C wcmtr 28

This theorem is referenced by:  wcom3ii 401

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-wom 343

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-i2 44  df-le 121  df-le1 122  df-le2 123  df-cmtr 126

metamath.org