Certification Problem

Input (TPDB TRS_Standard/AProVE_04/IJCAR_12)

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 3 components.

• The 1^st component contains the pair

  quot#(s(x),s(y),z)	→	quot#(x,y,z)	(17)
  quot#(x,0,s(z))	→	div#(x,s(z))	(18)
  div#(x,y)	→	quot#(x,y,y)	(16)
  div#(div(x,y),z)	→	div#(x,times(y,z))	(20)

  1.1.1 Subterm Criterion Processor

  We use the projection

  π(quot#)	=	1
  π(div#)	=	1

  and remove the pairs:

  quot#(s(x),s(y),z)	→	quot#(x,y,z)	(17)
  div#(div(x,y),z)	→	div#(x,times(y,z))	(20)

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the arctic semiring over the integers

  [div#(x1, x2)]	=	-∞ · x1 + 0 · x2 + 0
  [0]	=	1
  [quot#(x1, x2, x3)]	=	-∞ · x1 + 0 · x2 + 0 · x3 + 0
  [s(x1)]	=	-∞ · x1 + 0

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

  quot#(x,0,s(z))

  →

  div#(x,s(z))

  (18)

  could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  times#(s(x),y)

  →

  times#(x,y)

  (14)

  1.1.2 Size-Change Termination

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

  times#(s(x),y)	→	times#(x,y)	(14)
  2	≥	2
  1	>	1

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

• The 3^rd component contains the pair

  plus#(s(x),y)

  →

  plus#(x,y)

  (13)

  1.1.3 Size-Change Termination

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

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

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