Certification Problem

Input (TPDB TRS_Standard/SK90/2.06)

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

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [+(x1, x2)]	=	x1 + x2 + 1
  [+#(x1, x2)]	=	x2 + 0
  [*(x1, x2)]	=	x2 + 23676

  together with the usable rules

  +(x,+(y,z))	→	+(+(x,y),z)	(1)
  +(*(x,y),+(*(x,z),u))	→	+(*(x,+(y,z)),u)	(3)
  +(*(x,y),+(x,z))	→	*(x,+(y,z))	(2)

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

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

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.