Certification Problem

Input (TPDB TRS_Standard/SK90/4.17)

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 2 components.

• The 1^st component contains the pair

  fac#(s(x))

  →

  fac#(p(s(x)))

  (4)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [s(x1)]	=	0 1 0 1 · x1 + 15943 15944
  [fac#(x1)]	=	0 1 0 0 · x1 + 0 0
  [p#(x1)]	=	0 0
  [p(x1)]	=	1 1 1 0 · x1 + 0 0
  [0]	=	1 18008
  [fac(x1)]	=	0 0
  [*(x1, x2)]	=	0 0

  together with the usable rules

  p(s(s(x)))	→	s(p(s(x)))	(3)
  p(s(0))	→	0	(2)

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

  fac#(s(x))

  →

  fac#(p(s(x)))

  (4)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  p#(s(s(x)))

  →

  p#(s(x))

  (6)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [fac#(x1)]	=	0
  [p#(x1)]	=	x1 + 0
  [p(x1)]	=	11293
  [0]	=	11294
  [fac(x1)]	=	0
  [*(x1, x2)]	=	0

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

  p#(s(s(x)))

  →

  p#(s(x))

  (6)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.