Theorem wdid0id3 1114

Description: Show a quantum identity that follows from classical identity in a WDOL. (Contributed by NM, 5-Mar-2006.)

Hypothesis

Ref	Expression
wdid0id5.1	(a ≡0 b) = 1

Assertion

Ref	Expression
wdid0id3	(a ≡3 b) = 1

Proof of Theorem wdid0id3

Step	Hyp	Ref	Expression
1		df-id3 52	. 2 (a ≡3 b) = ((a⊥ ∪ b) ∩ (a ∪ (a⊥ ∩ b⊥ )))
2		df-id0 49	. . . . 5 (a ≡0 b) = ((a⊥ ∪ b) ∩ (b⊥ ∪ a))
3	2	ax-r1 35	. . . 4 ((a⊥ ∪ b) ∩ (b⊥ ∪ a)) = (a ≡0 b)
4		wdid0id5.1	. . . 4 (a ≡0 b) = 1
5	3, 4	ax-r2 36	. . 3 ((a⊥ ∪ b) ∩ (b⊥ ∪ a)) = 1
6		wa4 194	. . . . . . . 8 (((a ∪ b⊥ ) ∪ (a ∪ a⊥ )) ≡ (a ∪ a⊥ )) = 1
7	6	wleoa 376	. . . . . . 7 (((a ∪ b⊥ ) ∩ (a ∪ a⊥ )) ≡ (a ∪ b⊥ )) = 1
8	7	wr1 197	. . . . . 6 ((a ∪ b⊥ ) ≡ ((a ∪ b⊥ ) ∩ (a ∪ a⊥ ))) = 1
9		wancom 203	. . . . . 6 (((a ∪ b⊥ ) ∩ (a ∪ a⊥ )) ≡ ((a ∪ a⊥ ) ∩ (a ∪ b⊥ ))) = 1
10	8, 9	wr2 371	. . . . 5 ((a ∪ b⊥ ) ≡ ((a ∪ a⊥ ) ∩ (a ∪ b⊥ ))) = 1
11		wa2 192	. . . . 5 ((b⊥ ∪ a) ≡ (a ∪ b⊥ )) = 1
12		wddi3 1109	. . . . 5 ((a ∪ (a⊥ ∩ b⊥ )) ≡ ((a ∪ a⊥ ) ∩ (a ∪ b⊥ ))) = 1
13	10, 11, 12	w3tr1 374	. . . 4 ((b⊥ ∪ a) ≡ (a ∪ (a⊥ ∩ b⊥ ))) = 1
14	13	wlan 370	. . 3 (((a⊥ ∪ b) ∩ (b⊥ ∪ a)) ≡ ((a⊥ ∪ b) ∩ (a ∪ (a⊥ ∩ b⊥ )))) = 1
15	5, 14	wwbmp 205	. 2 ((a⊥ ∪ b) ∩ (a ∪ (a⊥ ∩ b⊥ ))) = 1
16	1, 15	ax-r2 36	1 (a ≡3 b) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   ≡₀ wid0 17   ≡₃ wid3 20

This theorem was proved from 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-wom 361  ax-wdol 1104

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-id0 49  df-id3 52  df-le 129  df-le1 130  df-le2 131  df-cmtr 134

This theorem is referenced by:  wddi-3  1120