Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.23)

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

  quot#(s(x),s(y),z)	→	quot#(x,y,z)	(8)
  quot#(x,0,s(z))	→	quot#(x,plus(z,s(0)),s(z))	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [plus#(x1, x2)]	=	0
  [0]	=	1
  [quot(x1, x2, x3)]	=	0
  [plus(x1, x2)]	=	x1 + 32286
  [quot#(x1, x2, x3)]	=	x1 + 0

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

  quot#(s(x),s(y),z)

  →

  quot#(x,y,z)

  (8)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    quot#(x,0,s(z))

    →

    quot#(x,plus(z,s(0)),s(z))

    (6)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	1
    [plus#(x1, x2)]	=	0
    [0]	=	3
    [quot(x1, x2, x3)]	=	0
    [plus(x1, x2)]	=	x2 + 1
    [quot#(x1, x2, x3)]	=	x2 + 0

    together with the usable rules

    plus(s(x),y)	→	s(plus(x,y))	(4)
    plus(0,y)	→	y	(3)

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

    quot#(x,0,s(z))

    →

    quot#(x,plus(z,s(0)),s(z))

    (6)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  plus#(s(x),y)

  →

  plus#(x,y)

  (9)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [plus#(x1, x2)]	=	x1 + 0
  [0]	=	20819
  [quot(x1, x2, x3)]	=	0
  [plus(x1, x2)]	=	x1 + x2 + 0
  [quot#(x1, x2, x3)]	=	x2 + 0

  together with the usable rules

  plus(s(x),y)	→	s(plus(x,y))	(4)
  plus(0,y)	→	y	(3)

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

  plus#(s(x),y)

  →

  plus#(x,y)

  (9)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.