Certification Problem

Input (TPDB TRS_Standard/Various_04/12)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  *#(I(x),y)	→	*#(x,y)	(21)
  *#(O(x),y)	→	*#(x,y)	(19)

  1.1.1 Subterm Criterion Processor

  We use the projection

  π(*#)

  =

  1

  and remove the pairs:

  *#(I(x),y)	→	*#(x,y)	(21)
  *#(O(x),y)	→	*#(x,y)	(19)

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  +#(O(x),I(y))	→	+#(x,y)	(14)
  +#(I(x),I(y))	→	+#(+(x,y),I(0))	(17)
  +#(I(x),I(y))	→	+#(x,y)	(16)
  +#(I(x),O(y))	→	+#(x,y)	(15)
  +#(O(x),O(y))	→	+#(x,y)	(12)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the arctic semiring over the integers

  [O(x1)]	=	1 · x1 + 7
  [I(x1)]	=	0 · x1 + -8
  [0]	=	6
  [+#(x1, x2)]	=	-∞ · x1 + 0 · x2 + 6
  [+(x1, x2)]	=	0 · x1 + 0 · x2 + 0

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

  +#(I(x),O(y))	→	+#(x,y)	(15)
  +#(O(x),O(y))	→	+#(x,y)	(12)

  could be deleted.

  1.1.2.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the arctic semiring over the integers

  [O(x1)]	=	0 · x1 + 2
  [I(x1)]	=	1 · x1 + 1
  [0]	=	0
  [+#(x1, x2)]	=	-∞ · x1 + 0 · x2 + 0
  [+(x1, x2)]	=	3 · x1 + 1 · x2 + -∞

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

  +#(O(x),I(y))	→	+#(x,y)	(14)
  +#(I(x),I(y))	→	+#(x,y)	(16)

  could be deleted.

  1.1.2.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [O(x1)]	=	1 · x1 + 1
  [I(x1)]	=	1 · x1 + 1
  [0]	=	0
  [+#(x1, x2)]	=	1 · x1 + 4 · x2 + 4
  [+(x1, x2)]	=	1 · x1 + 4 · x2 + 0

  together with the usable rules

  +(0,x)	→	x	(2)
  +(x,0)	→	x	(3)
  +(O(x),O(y))	→	O(+(x,y))	(4)
  +(O(x),I(y))	→	I(+(x,y))	(5)
  +(I(x),O(y))	→	I(+(x,y))	(6)
  +(I(x),I(y))	→	O(+(+(x,y),I(0)))	(7)
  O(0)	→	0	(1)

  (w.r.t. the implicit argument filter of the reduction pair), the pair

  +#(I(x),I(y))

  →

  +#(+(x,y),I(0))

  (17)

  could be deleted.

  1.1.2.1.1.1 P is empty

  There are no pairs anymore.