Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z26)

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(x,y)))	→	f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))	(13)
  f#(a(x),a(y))	→	a#(f(x,y))	(11)
  f#(b(x),b(y))	→	f#(x,y)	(10)
  a#(a(f(x,y)))	→	a#(y)	(8)
  a#(a(f(x,y)))	→	a#(x)	(7)
  f#(a(x),a(y))	→	f#(x,y)	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

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

  together with the usable rules

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

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

  a#(a(f(x,y)))	→	a#(y)	(8)
  a#(a(f(x,y)))	→	a#(x)	(7)

  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(x,y)))	→	f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))	(13)
    f#(b(x),b(y))	→	f#(x,y)	(10)
    f#(a(x),a(y))	→	f#(x,y)	(4)
    f#(a(x),a(y))	→	a#(f(x,y))	(11)

    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)]	=	1 1 1 0 · x1 + 0 1
    [b(x1)]	=	1 0 0 0 · x1 + 0 0
    [f(x1, x2)]	=	x1 + x2 + 1 0
    [f#(x1, x2)]	=	1 0 1 0 · x1 + 1 0 1 0 · x2 + 2 1
    [a#(x1)]	=	1 1 1 0 · x1 + 0 0

    together with the usable rules

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

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

    a#(a(f(x,y)))	→	f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))	(13)
    f#(a(x),a(y))	→	a#(f(x,y))	(11)

    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

      f#(b(x),b(y))	→	f#(x,y)	(10)
      f#(a(x),a(y))	→	f#(x,y)	(4)

      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)]	=	1 1 0 0 · x1 + 1 31584
      [b(x1)]	=	1 0 0 0 · x1 + 1 0
      [f(x1, x2)]	=	x1 + 22026 0
      [f#(x1, x2)]	=	1 0 1 0 · x2 + 2 1
      [a#(x1)]	=	0 0

      together with the usable rules

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

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

      f#(b(x),b(y))	→	f#(x,y)	(10)
      f#(a(x),a(y))	→	f#(x,y)	(4)

      could be deleted.

      1.1.1.1.1.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 0 components.