Theorem u1lemnonb 675

Description: Lemma for Sasaki implication study. (Contributed by NM, 16-Dec-1997.)

Assertion

Ref	Expression
u1lemnonb	((a →1 b)⊥ ∪ b⊥ ) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))

Proof of Theorem u1lemnonb

Step	Hyp	Ref	Expression
1		u1lemab 610	. . . 4 ((a →1 b) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))
2		ax-a2 31	. . . . 5 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a ∩ b))
3		anor2 89	. . . . . 6 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
4		df-a 40	. . . . . 6 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	3, 4	2or 72	. . . . 5 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
6	2, 5	ax-r2 36	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
7	1, 6	ax-r2 36	. . 3 ((a →1 b) ∩ b) = ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
8		df-a 40	. . 3 ((a →1 b) ∩ b) = ((a →1 b)⊥ ∪ b⊥ )⊥
9		oran3 93	. . 3 ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ ) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
10	7, 8, 9	3tr2 64	. 2 ((a →1 b)⊥ ∪ b⊥ )⊥ = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
11	10	con1 66	1 ((a →1 b)⊥ ∪ b⊥ ) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

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-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by: (None)