Certification Problem

Input (TPDB TRS_Standard/AG01/#3.22)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ 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,s(z))	(10)
  times#(x,s(y))	→	times#(x,y)	(12)
  times#(x,plus(y,s(z)))	→	times#(x,plus(y,times(s(z),0)))	(7)

  1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  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 rule could be deleted.

  1.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pair

  times#(x,plus(y,s(z)))

  →

  times#(x,s(z))

  (10)

  and the rule

  plus(x,0)

  →

  x

  (4)

  could be deleted.

  1.1.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pair

  times#(x,s(y))

  →

  times#(x,y)

  (12)

  and the rule

  plus(x,s(y))

  →

  s(plus(x,y))

  (5)

  could be deleted.

  1.1.1.1.1.1 Switch to Innermost Termination

  The TRS does not have overlaps with the pairs and is locally confluent:

  20

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

  1.1.1.1.1.1.1 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,s(z)))	→	times#(x,plus(y,times(s(z),0)))	(7)
  1	≥	1

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

• The 2^nd component contains the pair

  plus#(x,s(y))

  →

  plus#(x,y)

  (13)

  1.1.2 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [s(x1)]	=	1 · x1
  [plus#(x1, x2)]	=	1 · x1 + 1 · x2

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

  1.1.2.1 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)
  1	≥	1
  2	>	2

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