Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/direct)

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

  h#(c(s(x),c(s(0),y)),z)	→	h#(y,c(s(0),c(x,z)))	(4)
  h#(x,c(y,z))	→	h#(c(s(y),x),z)	(3)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [h(x1, x2)]	=	0 0
  [s(x1)]	=	0 1 1 0 · x1 + 32002 51021
  [c(x1, x2)]	=	x1 + x2 + 83746 32724
  [0]	=	32006 1
  [h#(x1, x2)]	=	0 1 0 0 · x1 + 1 0 0 0 · x2 + 0 0

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

  h#(c(s(x),c(s(0),y)),z)	→	h#(y,c(s(0),c(x,z)))	(4)
  h#(x,c(y,z))	→	h#(c(s(y),x),z)	(3)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.