Certification Problem

Input (TPDB TRS_Standard/AProVE_07/kabasci03)

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#(x,c(y,z),t(w))	→	h#(c(s(y),x),z,t(c(t(w),w)))	(13)
  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)
  h#(c(x,y),c(s(z),z),t(w))	→	h#(z,c(y,x),t(t(c(x,c(y,t(w))))))	(8)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [h(x1, x2, x3)]	=	0
  [s(x1)]	=	x1 + 0
  [t(x1)]	=	1
  [c(x1, x2)]	=	x1 + x2 + 5854
  [0]	=	29243
  [h#(x1, x2, x3)]	=	x1 + x2 + 0
  [t#(x1)]	=	0

  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))))))

  (8)

  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

    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)
    h#(x,c(y,z),t(w))	→	h#(c(s(y),x),z,t(c(t(w),w)))	(13)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

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

    [h(x1, x2, x3)]	=	0 0
    [s(x1)]	=	0 1 0 0 · x1 + 1 26229
    [t(x1)]	=	x1 + 326625 0
    [c(x1, x2)]	=	x1 + x2 + 24634 24636
    [0]	=	40691 26232
    [h#(x1, x2, x3)]	=	1 0 0 0 · x1 + 0 1 0 0 · x2 + 0 0
    [t#(x1)]	=	0 0

    together with the usable rule

    t(x)

    →

    x

    (5)

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

    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)
    h#(x,c(y,z),t(w))	→	h#(c(s(y),x),z,t(c(t(w),w)))	(13)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.