Theorem wdcom 1105

Description: Any two variables (weakly) commute in a WDOL. (Contributed by NM, 4-Mar-2006.)

Assertion

Ref	Expression
wdcom	C (a, b) = 1

Proof of Theorem wdcom

Step	Hyp	Ref	Expression
1		df-cmtr 134	. 2 C (a, b) = (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
2		or42 85	. 2 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ b⊥ ) ∪ (a⊥ ∩ b)))
3		dfb 94	. . . . 5 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
4		dfb 94	. . . . . 6 (a ≡ b⊥ ) = ((a ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))
5		ax-a1 30	. . . . . . . . 9 b = b⊥ ⊥
6	5	lan 77	. . . . . . . 8 (a⊥ ∩ b) = (a⊥ ∩ b⊥ ⊥ )
7	6	ax-r1 35	. . . . . . 7 (a⊥ ∩ b⊥ ⊥ ) = (a⊥ ∩ b)
8	7	lor 70	. . . . . 6 ((a ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ )) = ((a ∩ b⊥ ) ∪ (a⊥ ∩ b))
9	4, 8	ax-r2 36	. . . . 5 (a ≡ b⊥ ) = ((a ∩ b⊥ ) ∪ (a⊥ ∩ b))
10	3, 9	2or 72	. . . 4 ((a ≡ b) ∪ (a ≡ b⊥ )) = (((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ b⊥ ) ∪ (a⊥ ∩ b)))
11	10	ax-r1 35	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ b⊥ ) ∪ (a⊥ ∩ b))) = ((a ≡ b) ∪ (a ≡ b⊥ ))
12		ax-wdol 1104	. . 3 ((a ≡ b) ∪ (a ≡ b⊥ )) = 1
13	11, 12	ax-r2 36	. 2 (((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ b⊥ ) ∪ (a⊥ ∩ b))) = 1
14	1, 2, 13	3tr 65	1 C (a, b) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 8   C wcmtr 29

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-wdol 1104

This proof depends on definitions:  df-b 39  df-a 40  df-cmtr 134

This theorem is used by:  wdwom  1106  wddi1  1107  wddi3  1109