Certification Problem

Input (TPDB TRS_Standard/AProVE_07/kabasci03)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ 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#(x,c(y,z),t(w))	→	h#(c(s(y),x),z,t(c(t(w),w)))	(10)
  h#(c(x,y),c(s(z),z),t(w))	→	h#(z,c(y,x),t(t(c(x,c(y,t(w))))))	(7)
  h#(c(s(x),c(s(0),y)),z,t(x))	→	h#(y,c(s(0),c(x,z)),t(t(c(x,s(x)))))	(12)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [h#(x1, x2, x3)]	=	-2 + x1 + x2
  [t(x1)]	=	-2
  [c(x1, x2)]	=	2 + x1 + x2
  [0]	=	1
  [s(x1)]	=	x1

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

  h#(c(x,y),c(s(z),z),t(w))

  →

  h#(z,c(y,x),t(t(c(x,c(y,t(w))))))

  (7)

  could be deleted.

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the rationals with delta = 1/64

  [h#(x1, x2, x3)]	=	0 + 1/2 · x1 + 1/4 · x2 + 0 · x3
  [c(x1, x2)]	=	0 + 1/4 · x1 + 1 · x2
  [t(x1)]	=	0 + 0 · x1
  [s(x1)]	=	0 + 1/2 · x1
  [0]	=	1/2

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

  h#(c(s(x),c(s(0),y)),z,t(x))

  →

  h#(y,c(s(0),c(x,z)),t(t(c(x,s(x)))))

  (12)

  could be deleted.

  1.1.1.1.1 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  h#(x,c(y,z),t(w))	→	h#(c(s(y),x),z,t(c(t(w),w)))	(10)
  2	>	2

  As there is no critical graph in the transitive closure, there are no infinite chains.