Certification Problem

Input (TPDB TRS_Standard/SK90/4.31)

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 2 components.

• The 1^st component contains the pair

  purge#(.(x,y))

  →

  purge#(remove(x,y))

  (7)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [purge#(x1)]	=	x1 + 0
  [purge(x1)]	=	0
  [if(x1, x2, x3)]	=	x1 + x2 + 0
  [=(x1, x2)]	=	1
  [nil]	=	16575
  [.(x1, x2)]	=	x1 + x2 + 23677
  [remove#(x1, x2)]	=	0
  [remove(x1, x2)]	=	x1 + x2 + 23676

  together with the usable rules

  remove(x,.(y,z))	→	if(=(x,y),remove(x,z),.(y,remove(x,z)))	(4)
  remove(x,nil)	→	nil	(3)

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

  purge#(.(x,y))

  →

  purge#(remove(x,y))

  (7)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  remove#(x,.(y,z))	→	remove#(x,z)	(6)
  remove#(x,.(y,z))	→	remove#(x,z)	(6)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [purge#(x1)]	=	x1 + 0
  [purge(x1)]	=	0
  [if(x1, x2, x3)]	=	x1 + x2 + 0
  [=(x1, x2)]	=	1
  [nil]	=	16575
  [.(x1, x2)]	=	x1 + x2 + 23677
  [remove#(x1, x2)]	=	x2 + 0
  [remove(x1, x2)]	=	x1 + x2 + 20163

  together with the usable rules

  remove(x,.(y,z))	→	if(=(x,y),remove(x,z),.(y,remove(x,z)))	(4)
  remove(x,nil)	→	nil	(3)

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

  remove#(x,.(y,z))	→	remove#(x,z)	(6)
  remove#(x,.(y,z))	→	remove#(x,z)	(6)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.