Theorem df2c1 450

Description: Dual 'commutes' analogue for ≡ analogue of =.

Hypothesis

Ref	Expression
df2c1.1	a = ((a ∪ b) ∩ (a ∪ b⊥ ))

Assertion

Ref	Expression
df2c1	a C b

Proof of Theorem df2c1

Step	Hyp	Ref	Expression
1		df2c1.1	. . . . 5 a = ((a ∪ b) ∩ (a ∪ b⊥ ))
2		df-a 39	. . . . . 6 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )⊥
3		anor3 82	. . . . . . . . 9 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
4		anor3 82	. . . . . . . . 9 (a⊥ ∩ b⊥ ⊥ ) = (a ∪ b⊥ )⊥
5	3, 4	2or 67	. . . . . . . 8 ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ )) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )
6	5	ax-r1 34	. . . . . . 7 ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ ) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))
7	6	ax-r4 36	. . . . . 6 ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )⊥ = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥
8	2, 7	ax-r2 35	. . . . 5 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥
9	1, 8	ax-r2 35	. . . 4 a = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))⊥
10	9	con2 64	. . 3 a⊥ = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ ⊥ ))
11	10	df-c1 124	. 2 a⊥ C b⊥
12	11	comcom5 440	1 a C b

Syntax hints:   = wb 1   C wc 3  ^⊥ wn 4   ∪ wo 6   ∩ wa 7

This theorem is referenced by:  gsth 471

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  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-le1 122  df-le2 123  df-c1 124  df-c2 125

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