Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/linear1)

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

  a#(a(f(b,a(x))))	→	a#(a(a(x)))	(5)
  a#(a(f(b,a(x))))	→	a#(a(x))	(6)
  a#(a(x))	→	a#(f(a(x),b))	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 11798
  [b]	=	0
  [f(x1, x2)]	=	x1 + x2 + 0
  [f#(x1, x2)]	=	0
  [a#(x1)]	=	x1 + 0

  together with the usable rules

  a(a(f(b,a(x))))	→	f(a(a(a(x))),b)	(1)
  f(a(x),b)	→	f(b,a(x))	(3)
  a(a(x))	→	f(b,a(f(a(x),b)))	(2)

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

  a#(a(f(b,a(x))))

  →

  a#(a(x))

  (6)

  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

    a#(a(f(b,a(x))))	→	a#(a(a(x)))	(5)
    a#(a(x))	→	a#(f(a(x),b))	(4)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a(x1)]	=	32287
    [b]	=	1
    [f(x1, x2)]	=	1
    [f#(x1, x2)]	=	0
    [a#(x1)]	=	x1 + 0

    together with the usable rules

    a(a(f(b,a(x))))	→	f(a(a(a(x))),b)	(1)
    f(a(x),b)	→	f(b,a(x))	(3)
    a(a(x))	→	f(b,a(f(a(x),b)))	(2)

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

    a#(a(x))

    →

    a#(f(a(x),b))

    (4)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      a#(a(f(b,a(x))))

      →

      a#(a(a(x)))

      (5)

      1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

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

      [a(x1)]	=	0 1 1 0 · x1 + 1941 40687
      [b]	=	1 1
      [f(x1, x2)]	=	1 0 0 0 · x1 + 1 0 0 0 · x2 + 0 40686
      [f#(x1, x2)]	=	0 0
      [a#(x1)]	=	0 1 0 0 · x1 + 0 0

      together with the usable rules

      a(a(f(b,a(x))))	→	f(a(a(a(x))),b)	(1)
      f(a(x),b)	→	f(b,a(x))	(3)
      a(a(x))	→	f(b,a(f(a(x),b)))	(2)

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

      a#(a(f(b,a(x))))

      →

      a#(a(a(x)))

      (5)

      could be deleted.

      1.1.1.1.1.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 0 components.