Certification Problem

Input (TPDB TRS_Standard/AProVE_06/factorial1)

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),y)

  →

  fac#(p(s(x)),times(s(x),y))

  (18)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [times(x1, x2)]	=	0 · x1 + 0 · x2 + 0
  [plus(x1, x2)]	=	-1 · x1 + -∞ · x2 + 4
  [p(x1)]	=	-1 · x1 + 0
  [fac#(x1, x2)]	=	0 · x1 + -∞ · x2 + -16
  [0]	=	0
  [s(x1)]	=	2 · x1 + 1

  together with the usable rules

  p(s(0))	→	0	(5)
  p(s(s(x)))	→	s(p(s(x)))	(6)

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

  fac#(s(x),y)

  →

  fac#(p(s(x)),times(s(x),y))

  (18)

  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#(p(s(x)),y)

  (13)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

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

  together with the usable rules

  p(s(0))	→	0	(5)
  p(s(s(x)))	→	s(p(s(x)))	(6)

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

  times#(s(x),y)

  →

  times#(p(s(x)),y)

  (13)

  could be deleted.

  1.1.2.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  plus#(s(x),y)

  →

  plus#(p(s(x)),y)

  (11)

  1.1.3 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

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

  together with the usable rules

  p(s(0))	→	0	(5)
  p(s(s(x)))	→	s(p(s(x)))	(6)

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

  plus#(s(x),y)

  →

  plus#(p(s(x)),y)

  (11)

  could be deleted.

  1.1.3.1 P is empty

  There are no pairs anymore.

• The 4^th component contains the pair

  p#(s(s(x)))

  →

  p#(s(x))

  (15)

  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))	(15)
  1	>	1

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