Certification Problem

Input (TPDB TRS_Standard/SK90/2.16)

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

  g#(s(x),y)	→	g#(x,s(+(y,x)))	(11)
  g#(s(x),y)	→	g#(x,+(y,s(x)))	(10)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [1]	=	0
  [s(x1)]	=	x1 + 1
  [f(x1)]	=	0
  [0]	=	1
  [f#(x1)]	=	0
  [g#(x1, x2)]	=	x1 + 0
  [+(x1, x2)]	=	1
  [g(x1, x2)]	=	0
  [+#(x1, x2)]	=	0

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

  g#(s(x),y)	→	g#(x,s(+(y,x)))	(11)
  g#(s(x),y)	→	g#(x,+(y,s(x)))	(10)

  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

  +#(x,s(y))

  →

  +#(x,y)

  (8)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [1]	=	0
  [s(x1)]	=	x1 + 1
  [f(x1)]	=	0
  [0]	=	1
  [f#(x1)]	=	0
  [g#(x1, x2)]	=	0
  [+(x1, x2)]	=	1
  [g(x1, x2)]	=	0
  [+#(x1, x2)]	=	x2 + 0

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

  +#(x,s(y))

  →

  +#(x,y)

  (8)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.