Certification Problem

Input (TPDB TRS_Standard/SK90/2.02)

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)	→	+#(x,+(y,z))	(8)
  +#(f(x),f(y))	→	+#(x,y)	(7)
  +#(f(x),+(f(y),z))	→	+#(x,y)	(6)
  +#(+(x,y),z)	→	+#(y,z)	(5)
  +#(f(x),+(f(y),z))	→	+#(f(+(x,y)),z)	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [f(x1)]	=	x1 + 1
  [+(x1, x2)]	=	x1 + x2 + 39
  [+#(x1, x2)]	=	x1 + x2 + 0

  together with the usable rules

  +(+(x,y),z)	→	+(x,+(y,z))	(1)
  +(f(x),+(f(y),z))	→	+(f(+(x,y)),z)	(3)
  +(f(x),f(y))	→	f(+(x,y))	(2)

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

  +#(f(x),f(y))	→	+#(x,y)	(7)
  +#(f(x),+(f(y),z))	→	+#(x,y)	(6)
  +#(+(x,y),z)	→	+#(y,z)	(5)
  +#(f(x),+(f(y),z))	→	+#(f(+(x,y)),z)	(4)

  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))

    (8)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [f(x1)]	=	x1 + 1
    [+(x1, x2)]	=	x1 + x2 + 39
    [+#(x1, x2)]	=	x1 + 0

    together with the usable rules

    +(+(x,y),z)	→	+(x,+(y,z))	(1)
    +(f(x),+(f(y),z))	→	+(f(+(x,y)),z)	(3)
    +(f(x),f(y))	→	f(+(x,y))	(2)

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

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

    →

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

    (8)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.