Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/013)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  app#(app(flatwithsub,f),app(app(cons,x),xs))	→	app#(app(flatwithsub,f),xs)	(15)
  app#(app(flatwithsub,f),app(app(cons,x),xs))	→	app#(app(flatwith,f),x)	(17)
  app#(app(flatwith,f),app(node,xs))	→	app#(app(flatwithsub,f),xs)	(14)
  app#(app(flatwith,f),app(leaf,x))	→	app#(f,x)	(10)

  1.1.1 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  app#(app(flatwithsub,f),app(app(cons,x),xs))	→	app#(app(flatwithsub,f),xs)	(15)
  2	>	2
  1	≥	1
  app#(app(flatwithsub,f),app(app(cons,x),xs))	→	app#(app(flatwith,f),x)	(17)
  2	>	2
  app#(app(flatwith,f),app(node,xs))	→	app#(app(flatwithsub,f),xs)	(14)
  2	>	2
  app#(app(flatwith,f),app(leaf,x))	→	app#(f,x)	(10)
  2	>	2
  1	>	1

  As there is no critical graph in the transitive closure, there are no infinite chains.

• The 2^nd component contains the pair

  app#(app(append,app(app(cons,x),xs)),ys)

  →

  app#(app(append,xs),ys)

  (8)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the

  prec(app#)	=	0		stat(app#)	=	lex
  prec(cons)	=	0		stat(cons)	=	lex
  prec(app)	=	1		stat(app)	=	lex
  prec(append)	=	0		stat(append)	=	lex

  π(app#)	=	[1]
  π(cons)	=	[]
  π(app)	=	[2]
  π(append)	=	[]

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

  app#(app(append,app(app(cons,x),xs)),ys)

  →

  app#(app(append,xs),ys)

  (8)

  could be deleted.

  1.1.2.1 P is empty

  There are no pairs anymore.