Certification Problem

Input (TPDB TRS_Standard/AProVE_04/fac)

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 4 components.

• The 1^st component contains the pair

  fac#(s(x))

  →

  fac#(p(s(x)))

  (14)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [p(x1)]	=	-1 · x1 + 0
  [0]	=	4
  [fac#(x1)]	=	0 · x1 + 0
  [s(x1)]	=	1 · x1 + 1

  together with the usable rules

  p(s(s(x)))	→	s(p(s(x)))	(6)
  p(s(0))	→	0	(7)

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

  fac#(s(x))

  →

  fac#(p(s(x)))

  (14)

  could be deleted.

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  times#(s(x),y)

  →

  times#(x,y)

  (10)

  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)	(10)
  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#(x,s(y))

  →

  plus#(x,y)

  (9)

  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#(x,s(y))	→	plus#(x,y)	(9)
  2	>	2
  1	≥	1

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

• The 4^th component contains the pair

  p#(s(s(x)))

  →

  p#(s(x))

  (12)

  1.1.4 Size-Change Termination

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

  p#(s(s(x)))	→	p#(s(x))	(12)
  1	>	1

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