Certification Problem

Input (TPDB TRS_Standard/AG01/#3.22)

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

  times#(x,plus(y,s(z)))	→	times#(x,plus(y,times(s(z),0)))	(9)
  times#(x,s(y))	→	times#(x,y)	(11)
  times#(x,plus(y,s(z)))	→	times#(x,s(z))	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the

  prec(times#)	=	0		stat(times#)	=	lex
  prec(0)	=	0		stat(0)	=	lex
  prec(times)	=	0		stat(times)	=	lex
  prec(plus)	=	4		stat(plus)	=	lex
  prec(s)	=	2		stat(s)	=	lex

  π(times#)	=	2
  π(0)	=	[]
  π(times)	=	2
  π(plus)	=	[1,2]
  π(s)	=	[1]

  together with the usable rules

  times(x,0)	→	0	(2)
  plus(x,0)	→	x	(4)
  plus(x,s(y))	→	s(plus(x,y))	(5)

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

  times#(x,plus(y,s(z)))	→	times#(x,plus(y,times(s(z),0)))	(9)
  times#(x,s(y))	→	times#(x,y)	(11)
  times#(x,plus(y,s(z)))	→	times#(x,s(z))	(6)

  could be deleted.

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  plus#(x,s(y))

  →

  plus#(x,y)

  (13)

  1.1.2 Size-Change Termination

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

  plus#(x,s(y))	→	plus#(x,y)	(13)
  2	>	2
  1	≥	1

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