Certification Problem

Input (TPDB TRS_Standard/SK90/4.48)

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

  f#(f(x,y,z),u,f(x,y,v))	→	f#(x,y,f(z,u,v))	(7)
  f#(f(x,y,z),u,f(x,y,v))	→	f#(z,u,v)	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [f(x1, x2, x3)]	=	x1 + x2 + x3 + 1
  [f#(x1, x2, x3)]	=	x1 + x2 + x3 + 0
  [g(x1)]	=	23676

  together with the usable rules

  f(x,x,y)	→	x	(4)
  f(f(x,y,z),u,f(x,y,v))	→	f(x,y,f(z,u,v))	(1)
  f(x,y,g(y))	→	x	(3)
  f(g(x),x,y)	→	y	(5)
  f(x,y,y)	→	y	(2)

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

  f#(f(x,y,z),u,f(x,y,v))	→	f#(x,y,f(z,u,v))	(7)
  f#(f(x,y,z),u,f(x,y,v))	→	f#(z,u,v)	(6)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.