Certification Problem

Input (TPDB TRS_Standard/AG01/#3.16)

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

  times#(x,s(y))

  →

  times#(x,y)

  (7)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

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

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

  times#(x,s(y))

  →

  times#(x,y)

  (7)

  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

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

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

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

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

  plus#(s(x),y)

  →

  plus#(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

    plus#(x,s(y))

    →

    plus#(x,y)

    (10)

    1.1.2.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

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

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

    plus#(x,s(y))

    →

    plus#(x,y)

    (10)

    could be deleted.

    1.1.2.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.