Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/07)

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(a(x1)))	→	a#(b(b(x1)))	(11)
  b#(a(x1))	→	b#(x1)	(10)
  b#(a(b(x1)))	→	b#(a(a(x1)))	(9)
  b#(a(b(x1)))	→	a#(x1)	(8)
  a#(a(a(x1)))	→	b#(x1)	(5)
  a#(a(a(x1)))	→	b#(b(x1))	(7)
  b#(a(x1))	→	b#(b(x1))	(6)
  b#(a(b(x1)))	→	a#(a(x1))	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 2438
  [b(x1)]	=	x1 + 2438
  [a#(x1)]	=	x1 + 0
  [b#(x1)]	=	x1 + 2437

  together with the usable rules

  b(a(x1))	→	b(b(x1))	(1)
  a(a(a(x1)))	→	a(b(b(x1)))	(3)
  b(a(b(x1)))	→	b(a(a(x1)))	(2)

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

  b#(a(x1))	→	b#(x1)	(10)
  b#(a(b(x1)))	→	a#(x1)	(8)
  a#(a(a(x1)))	→	b#(x1)	(5)
  a#(a(a(x1)))	→	b#(b(x1))	(7)
  b#(a(b(x1)))	→	a#(a(x1))	(4)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 2 components.

  • The 1^st component contains the pair

    a#(a(a(x1)))

    →

    a#(b(b(x1)))

    (11)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a(x1)]	=	x1 + 40651
    [b(x1)]	=	81301
    [a#(x1)]	=	x1 + 0
    [b#(x1)]	=	x1 + 2437

    together with the usable rules

    b(a(x1))	→	b(b(x1))	(1)
    a(a(a(x1)))	→	a(b(b(x1)))	(3)
    b(a(b(x1)))	→	b(a(a(x1)))	(2)

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

    a#(a(a(x1)))

    →

    a#(b(b(x1)))

    (11)

    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

    b#(a(x1))	→	b#(b(x1))	(6)
    b#(a(b(x1)))	→	b#(a(a(x1)))	(9)

    1.1.1.1.2 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a(x1)]	=	30613
    [b(x1)]	=	0
    [a#(x1)]	=	x1 + 0
    [b#(x1)]	=	x1 + 2437

    together with the usable rules

    b(a(x1))	→	b(b(x1))	(1)
    a(a(a(x1)))	→	a(b(b(x1)))	(3)
    b(a(b(x1)))	→	b(a(a(x1)))	(2)

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

    b#(a(x1))

    →

    b#(b(x1))

    (6)

    could be deleted.

    1.1.1.1.2.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      b#(a(b(x1)))

      →

      b#(a(a(x1)))

      (9)

      1.1.1.1.2.1.1 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 + 30809 1
      [b(x1)]	=	0 0 1 0 · x1 + 1 30809
      [a#(x1)]	=	0 0
      [b#(x1)]	=	1 0 0 0 · x1 + 0 0

      together with the usable rules

      b(a(x1))	→	b(b(x1))	(1)
      a(a(a(x1)))	→	a(b(b(x1)))	(3)
      b(a(b(x1)))	→	b(a(a(x1)))	(2)

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

      b#(a(b(x1)))

      →

      b#(a(a(x1)))

      (9)

      could be deleted.

      1.1.1.1.2.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 0 components.