Certification Problem

Input (TPDB TRS_Standard/SK90/2.38)

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

  ++#(.(x,y),z)	→	++#(y,z)	(7)
  ++#(++(x,y),z)	→	++#(y,z)	(6)
  ++#(++(x,y),z)	→	++#(x,++(y,z))	(5)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [++(x1, x2)]	=	x1 + x2 + 1
  [++#(x1, x2)]	=	x1 + x2 + 0
  [nil]	=	14235
  [.(x1, x2)]	=	x2 + 8856

  together with the usable rules

  ++(++(x,y),z)	→	++(x,++(y,z))	(4)
  ++(nil,y)	→	y	(1)
  ++(.(x,y),z)	→	.(x,++(y,z))	(3)
  ++(x,nil)	→	x	(2)

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

  ++#(.(x,y),z)	→	++#(y,z)	(7)
  ++#(++(x,y),z)	→	++#(y,z)	(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

    ++#(++(x,y),z)

    →

    ++#(x,++(y,z))

    (5)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [++(x1, x2)]	=	x1 + x2 + 1
    [++#(x1, x2)]	=	x1 + 0
    [nil]	=	36466
    [.(x1, x2)]	=	x2 + 8856

    together with the usable rule

    ++(x,nil)

    →

    x

    (2)

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

    ++#(++(x,y),z)

    →

    ++#(x,++(y,z))

    (5)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.