Certification Problem

Input (TPDB TRS_Standard/SK90/2.45)

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

  admit#(x,.(u,.(v,.(w,z))))

  →

  admit#(carry(x,u,v),z)

  (4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [admit(x1, x2)]	=	0
  [w]	=	1
  [true]	=	0
  [sum(x1, x2, x3)]	=	0
  [cond#(x1, x2)]	=	0
  [=(x1, x2)]	=	0
  [nil]	=	0
  [.(x1, x2)]	=	x2 + 23676
  [admit#(x1, x2)]	=	x1 + x2 + 0
  [carry(x1, x2, x3)]	=	1
  [cond(x1, x2)]	=	0

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

  admit#(x,.(u,.(v,.(w,z))))

  →

  admit#(carry(x,u,v),z)

  (4)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.