Certification Problem

Input (TPDB TRS_Standard/Applicative_05/TypeEx1)

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 1 component.

• The 1^st component contains the pair

  ap#(ap(ff,x),x)	→	ap#(ap(x,ap(ff,x)),ap(ap(cons,x),nil))	(4)
  ap#(ap(ff,x),x)	→	ap#(x,ap(ff,x))	(2)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

  [ap(x1, x2)]	=	0 1 0 0 · x1 + 0 1 0 0 · x2 + 1 0
  [ff]	=	1 2
  [nil]	=	1 0
  [ap#(x1, x2)]	=	1 0 1 0 · x1 + 1 0 1 0 · x2 + 0 0
  [cons]	=	1 1

  together with the usable rule

  ap(ap(ff,x),x)

  →

  ap(ap(x,ap(ff,x)),ap(ap(cons,x),nil))

  (1)

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

  ap#(ap(ff,x),x)

  →

  ap#(ap(x,ap(ff,x)),ap(ap(cons,x),nil))

  (4)

  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

    ap#(ap(ff,x),x)

    →

    ap#(x,ap(ff,x))

    (2)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

    [ap(x1, x2)]	=	0 0 1 0 · x1 + 0 0 0 1 · x2 + 0 0
    [ff]	=	1 1
    [nil]	=	1 0
    [ap#(x1, x2)]	=	0 1 0 1 · x1 + 0 0
    [cons]	=	20694 1

    together with the usable rule

    ap(ap(ff,x),x)

    →

    ap(ap(x,ap(ff,x)),ap(ap(cons,x),nil))

    (1)

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

    ap#(ap(ff,x),x)

    →

    ap#(x,ap(ff,x))

    (2)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.