Certification Problem

Input (TPDB TRS_Standard/SK90/4.42)

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#(g(x),y,z)	→	h#(x,y,z)	(9)
  f#(g(x),a)	→	f#(x,g(a))	(8)
  h#(g(x),y,z)	→	f#(y,h(x,y,z))	(7)
  f#(g(x),g(y))	→	h#(g(y),x,g(y))	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a]	=	1
  [h(x1, x2, x3)]	=	x1 + 3
  [f(x1, x2)]	=	x1 + 2
  [h#(x1, x2, x3)]	=	x2 + 1
  [f#(x1, x2)]	=	x1 + 0
  [g(x1)]	=	x1 + 2

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

  f#(g(x),a)	→	f#(x,g(a))	(8)
  h#(g(x),y,z)	→	f#(y,h(x,y,z))	(7)
  f#(g(x),g(y))	→	h#(g(y),x,g(y))	(6)

  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#(g(x),y,z)

    →

    h#(x,y,z)

    (9)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a]	=	1
    [h(x1, x2, x3)]	=	x1 + 1
    [f(x1, x2)]	=	x1 + 0
    [h#(x1, x2, x3)]	=	x1 + 1
    [f#(x1, x2)]	=	0
    [g(x1)]	=	x1 + 1

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

    h#(g(x),y,z)

    →

    h#(x,y,z)

    (9)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.