Certification Problem

Input (TPDB TRS_Standard/HirokawaMiddeldorp_04/t003)

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 3 components.

• The 1^st component contains the pair

  f#(s(x),s(y),z,u)	→	f#(s(x),-(y,x),z,u)	(23)
  f#(s(x),0,z,u)	→	f#(x,u,-(z,s(x)),u)	(22)
  f#(s(x),s(y),z,u)	→	f#(x,u,z,u)	(15)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [<=(x1, x2)]	=	0
  [perfectp#(x1)]	=	0
  [false]	=	0
  [<=#(x1, x2)]	=	0
  [true]	=	0
  [f(x1,...,x4)]	=	0
  [0]	=	2
  [if(x1, x2, x3)]	=	0
  [-(x1, x2)]	=	x2 + 1
  [f#(x1,...,x4)]	=	x1 + 0
  [-#(x1, x2)]	=	0
  [if#(x1, x2, x3)]	=	0
  [perfectp(x1)]	=	0

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

  f#(s(x),0,z,u)	→	f#(x,u,-(z,s(x)),u)	(22)
  f#(s(x),s(y),z,u)	→	f#(x,u,z,u)	(15)

  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

    f#(s(x),s(y),z,u)

    →

    f#(s(x),-(y,x),z,u)

    (23)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 2
    [<=(x1, x2)]	=	0
    [perfectp#(x1)]	=	0
    [false]	=	0
    [<=#(x1, x2)]	=	0
    [true]	=	0
    [f(x1,...,x4)]	=	0
    [0]	=	2
    [if(x1, x2, x3)]	=	0
    [-(x1, x2)]	=	x1 + 1
    [f#(x1,...,x4)]	=	x2 + 0
    [-#(x1, x2)]	=	0
    [if#(x1, x2, x3)]	=	0
    [perfectp(x1)]	=	0

    together with the usable rules

    -(x,0)	→	x	(1)
    -(s(x),s(y))	→	-(x,y)	(2)

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

    f#(s(x),s(y),z,u)

    →

    f#(s(x),-(y,x),z,u)

    (23)

    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

  <=#(s(x),s(y))

  →

  <=#(x,y)

  (18)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 2
  [<=(x1, x2)]	=	0
  [perfectp#(x1)]	=	0
  [false]	=	0
  [<=#(x1, x2)]	=	x2 + 0
  [true]	=	0
  [f(x1,...,x4)]	=	0
  [0]	=	2
  [if(x1, x2, x3)]	=	0
  [-(x1, x2)]	=	x1 + 1
  [f#(x1,...,x4)]	=	x2 + 0
  [-#(x1, x2)]	=	0
  [if#(x1, x2, x3)]	=	0
  [perfectp(x1)]	=	0

  together with the usable rules

  -(x,0)	→	x	(1)
  -(s(x),s(y))	→	-(x,y)	(2)

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

  <=#(s(x),s(y))

  →

  <=#(x,y)

  (18)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 3^rd component contains the pair

  -#(s(x),s(y))

  →

  -#(x,y)

  (21)

  1.1.3 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 2
  [<=(x1, x2)]	=	0
  [perfectp#(x1)]	=	0
  [false]	=	0
  [<=#(x1, x2)]	=	0
  [true]	=	0
  [f(x1,...,x4)]	=	0
  [0]	=	2
  [if(x1, x2, x3)]	=	0
  [-(x1, x2)]	=	x1 + 1
  [f#(x1,...,x4)]	=	x2 + 0
  [-#(x1, x2)]	=	x2 + 0
  [if#(x1, x2, x3)]	=	0
  [perfectp(x1)]	=	0

  together with the usable rules

  -(x,0)	→	x	(1)
  -(s(x),s(y))	→	-(x,y)	(2)

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

  -#(s(x),s(y))

  →

  -#(x,y)

  (21)

  could be deleted.

  1.1.3.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.