Certification Problem

Input (TPDB TRS_Standard/SK90/2.52)

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 4 components.

• The 1^st component contains the pair

  f#(s(x),s(y),s(z))	→	f#(s(x),s(y),z)	(19)
  f#(s(x),0,s(z))	→	f#(x,s(0),z)	(18)
  f#(s(x),s(y),0)	→	f#(x,y,s(0))	(13)
  f#(s(x),s(y),s(z))	→	f#(x,y,f(s(x),s(y),z))	(15)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 2
  [f(x1, x2, x3)]	=	1
  [0]	=	1
  [f#(x1, x2, x3)]	=	x1 + 0

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

  f#(s(x),0,s(z))	→	f#(x,s(0),z)	(18)
  f#(s(x),s(y),0)	→	f#(x,y,s(0))	(13)
  f#(s(x),s(y),s(z))	→	f#(x,y,f(s(x),s(y),z))	(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),s(z))

    →

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

    (19)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 1
    [f(x1, x2, x3)]	=	max(0)
    [0]	=	0
    [f#(x1, x2, x3)]	=	max(x3 + 1, 0)

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

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

    →

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

    (19)

    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

  f#(0,s(s(y)),s(s(z)))	→	f#(0,s(s(y)),s(z))	(20)
  f#(0,s(s(y)),s(s(z)))	→	f#(0,y,f(0,s(s(y)),s(z)))	(17)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 2
  [f(x1, x2, x3)]	=	x2 + 0
  [0]	=	1
  [f#(x1, x2, x3)]	=	x2 + 0

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

  f#(0,s(s(y)),s(s(z)))

  →

  f#(0,y,f(0,s(s(y)),s(z)))

  (17)

  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

    f#(0,s(s(y)),s(s(z)))

    →

    f#(0,s(s(y)),s(z))

    (20)

    1.1.2.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 10452
    [f(x1, x2, x3)]	=	max(0)
    [0]	=	0
    [f#(x1, x2, x3)]	=	max(x1 + 20903, x3 + 0, 0)

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

    f#(0,s(s(y)),s(s(z)))

    →

    f#(0,s(s(y)),s(z))

    (20)

    could be deleted.

    1.1.2.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.

• The 3^rd component contains the pair

  f#(0,s(0),s(s(z)))

  →

  f#(0,s(0),z)

  (16)

  1.1.3 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 4
  [f(x1, x2, x3)]	=	max(0)
  [0]	=	1
  [f#(x1, x2, x3)]	=	max(x1 + 28106, x2 + 28100, x3 + 28101, 0)

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

  f#(0,s(0),s(s(z)))

  →

  f#(0,s(0),z)

  (16)

  could be deleted.

  1.1.3.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 4^th component contains the pair

  f#(0,s(s(y)),s(0))

  →

  f#(0,y,s(0))

  (12)

  1.1.4 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 2
  [f(x1, x2, x3)]	=	x2 + 0
  [0]	=	1
  [f#(x1, x2, x3)]	=	x2 + 0

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

  f#(0,s(s(y)),s(0))

  →

  f#(0,y,s(0))

  (12)

  could be deleted.

  1.1.4.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.