Certification Problem

Input (TPDB TRS_Standard/AG01/#3.21)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  times#(x,plus(y,1))

  →

  times#(x,plus(y,times(1,0)))

  (6)

  1.1.1.1 Usable Rules Processor

  We restrict the rewrite rules to the following usable rules of the DP problem.

  times(x,0)	→	0	(4)
  plus(x,0)	→	x	(3)

  1.1.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [times(x1, x2)]	=	2 · x1 + 1 · x2
  [0]	=	0
  [plus(x1, x2)]	=	1 + 1 · x1 + 1 · x2
  [times#(x1, x2)]	=	1 · x1 + 2 · x2
  [1]	=	0

  the rules

  times(x,plus(y,1))	→	plus(times(x,plus(y,times(1,0))),x)	(1)
  times(x,1)	→	x	(2)
  plus(x,0)	→	x	(3)

  could be deleted.

  1.1.1.1.1.1 Size-Change Termination

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

  times#(x,plus(y,1))	→	times#(x,plus(y,times(1,0)))	(6)
  1	≥	1

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