Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/027)

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#(inc,xs)	→	app#(app(map,app(plus,app(s,0))),xs)	(12)
  app#(app(map,f),app(app(cons,x),xs))	→	app#(f,x)	(14)
  app#(app(map,f),app(app(cons,x),xs))	→	app#(app(map,f),xs)	(13)

  1.1.1 Subterm Criterion Processor

  We use the projection

  π(app#)

  =

  2

  and remove the pairs:

  app#(app(map,f),app(app(cons,x),xs))	→	app#(f,x)	(14)
  app#(app(map,f),app(app(cons,x),xs))	→	app#(app(map,f),xs)	(13)

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  app#(app(plus,app(s,x)),y)

  →

  app#(app(plus,x),y)

  (7)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the

  prec(app#)	=	0		stat(app#)	=	lex
  prec(s)	=	0		stat(s)	=	lex
  prec(app)	=	1		stat(app)	=	lex
  prec(plus)	=	0		stat(plus)	=	lex

  π(app#)	=	[1]
  π(s)	=	[]
  π(app)	=	[2]
  π(plus)	=	[]

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

  app#(app(plus,app(s,x)),y)

  →

  app#(app(plus,x),y)

  (7)

  could be deleted.

  1.1.2.1 P is empty

  There are no pairs anymore.