Certification Problem

Input (TPDB TRS_Standard/SK90/2.13)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ 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

  double#(s(x))

  →

  double#(x)

  (9)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [0]	=	0
  [double#(x1)]	=	x1 + 0
  [double(x1)]	=	0
  [+(x1, x2)]	=	0
  [+#(x1, x2)]	=	0

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

  double#(s(x))

  →

  double#(x)

  (9)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  +#(x,s(y))	→	+#(x,y)	(10)
  +#(s(x),y)	→	+#(x,y)	(8)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [0]	=	0
  [double#(x1)]	=	0
  [double(x1)]	=	0
  [+(x1, x2)]	=	0
  [+#(x1, x2)]	=	x1 + 0

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

  +#(s(x),y)

  →

  +#(x,y)

  (8)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    +#(x,s(y))

    →

    +#(x,y)

    (10)

    1.1.2.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 1
    [0]	=	0
    [double#(x1)]	=	0
    [double(x1)]	=	0
    [+(x1, x2)]	=	0
    [+#(x1, x2)]	=	x2 + 0

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

    +#(x,s(y))

    →

    +#(x,y)

    (10)

    could be deleted.

    1.1.2.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.