Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z013)

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

  g#(g(x1))	→	f#(g(x1))	(12)
  f#(f(g(x1)))	→	f#(x1)	(11)
  g#(g(x1))	→	g#(f(g(x1)))	(9)
  f#(f(g(x1)))	→	g#(f(x1))	(7)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [c(x1)]	=	42736
  [f(x1)]	=	x1 + 0
  [f#(x1)]	=	x1 + 0
  [g#(x1)]	=	x1 + 8366
  [g(x1)]	=	x1 + 8367

  together with the usable rules

  f(f(g(x1)))	→	g(f(x1))	(4)
  g(c(x1))	→	g(f(c(x1)))	(1)
  g(g(x1))	→	g(f(g(x1)))	(3)
  g(f(c(x1)))	→	g(f(f(c(x1))))	(2)

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

  g#(g(x1))	→	f#(g(x1))	(12)
  f#(f(g(x1)))	→	f#(x1)	(11)
  f#(f(g(x1)))	→	g#(f(x1))	(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

    g#(g(x1))

    →

    g#(f(g(x1)))

    (9)

    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

    [c(x1)]	=	1 1 1 1 · x1 + 0 1
    [f(x1)]	=	0 1 1 0 · x1 + 3153 16674
    [f#(x1)]	=	0 0
    [g#(x1)]	=	0 1 0 0 · x1 + 0 0
    [g(x1)]	=	18005 34680

    together with the usable rules

    f(f(g(x1)))	→	g(f(x1))	(4)
    g(c(x1))	→	g(f(c(x1)))	(1)
    g(g(x1))	→	g(f(g(x1)))	(3)
    g(f(c(x1)))	→	g(f(f(c(x1))))	(2)

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

    g#(g(x1))

    →

    g#(f(g(x1)))

    (9)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.