Certification Problem

Input (TPDB TRS_Standard/AProVE_10/scnp)

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

  f#(x,s(y))	→	f#(y,x)	(6)
  f#(s(x),y)	→	f#(x,s(x))	(5)
  f#(c(x),y)	→	f#(x,s(x))	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [c(x1)]	=	x1 + 2
  [f(x1, x2)]	=	max(0)
  [f#(x1, x2)]	=	max(x1 + 28958, x2 + 28958, 0)

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

  f#(c(x),y)

  →

  f#(x,s(x))

  (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

    f#(s(x),y)	→	f#(x,s(x))	(5)
    f#(x,s(y))	→	f#(y,x)	(6)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 2
    [c(x1)]	=	2
    [f(x1, x2)]	=	max(0)
    [f#(x1, x2)]	=	max(x1 + 2, x2 + 1, 0)

    having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

    f#(s(x),y)	→	f#(x,s(x))	(5)
    f#(x,s(y))	→	f#(y,x)	(6)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.