Certification Problem

Input (TPDB TRS_Standard/AG01/#3.57)

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 6 components.

• The 1^st component contains the pair

  quot#(s(x),s(y))

  →

  quot#(minus(x,y),s(y))

  (16)

  1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(quot#)	=	{ 1 }
  π(minus)	=	{ 1 }

  to remove the pairs:

  quot#(s(x),s(y))

  →

  quot#(minus(x,y),s(y))

  (16)

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  minus#(minus(x,y),z)	→	minus#(x,plus(y,z))	(19)
  minus#(s(x),s(y))	→	minus#(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.

  minus#(minus(x,y),z)	→	minus#(x,plus(y,z))	(19)
  1	>	1
  minus#(s(x),s(y))	→	minus#(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

  sum#(app(l,cons(x,cons(y,k))))

  →

  sum#(app(l,sum(cons(x,cons(y,k)))))

  (25)

  1.1.3 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [plus(x1, x2)]	=	5 · x1 + 4 · x2 + 1
  [sum(x1)]	=	0 · x1 + 1
  [nil]	=	0
  [0]	=	2
  [cons(x1, x2)]	=	0 · x1 + 1 · x2 + 1
  [sum#(x1)]	=	4 · x1 + 4
  [app(x1, x2)]	=	4 · x1 + 2 · x2 + 0
  [s(x1)]	=	0 · x1 + 3

  together with the usable rules

  sum(cons(x,cons(y,l)))	→	sum(cons(plus(x,y),l))	(12)
  sum(cons(x,nil))	→	cons(x,nil)	(11)
  plus(0,y)	→	y	(5)
  plus(s(x),y)	→	s(plus(x,y))	(6)
  app(nil,k)	→	k	(8)
  app(l,nil)	→	l	(9)
  app(cons(x,l),k)	→	cons(x,app(l,k))	(10)

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

  sum#(app(l,cons(x,cons(y,k))))

  →

  sum#(app(l,sum(cons(x,cons(y,k)))))

  (25)

  could be deleted.

  1.1.3.1 P is empty

  There are no pairs anymore.

• The 4^th component contains the pair

  app#(cons(x,l),k)

  →

  app#(l,k)

  (20)

  1.1.4 Subterm Criterion Processor

  We use the projection

  π(app#)

  =

  1

  and remove the pairs:

  app#(cons(x,l),k)

  →

  app#(l,k)

  (20)

  1.1.4.1 P is empty

  There are no pairs anymore.

• The 5^th component contains the pair

  sum#(cons(x,cons(y,l)))

  →

  sum#(cons(plus(x,y),l))

  (22)

  1.1.5 Reduction Pair Processor with Usable Rules

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

  [plus(x1, x2)]	=	-∞ · x1 + 0 · x2 + 1
  [0]	=	2
  [cons(x1, x2)]	=	0 · x1 + 3 · x2 + 1
  [sum#(x1)]	=	0 · x1 + 0
  [s(x1)]	=	-∞ · x1 + 0

  together with the usable rules

  plus(0,y)	→	y	(5)
  plus(s(x),y)	→	s(plus(x,y))	(6)

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

  sum#(cons(x,cons(y,l)))

  →

  sum#(cons(plus(x,y),l))

  (22)

  could be deleted.

  1.1.5.1 P is empty

  There are no pairs anymore.

• The 6^th component contains the pair

  plus#(s(x),y)

  →

  plus#(x,y)

  (17)

  1.1.6 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)	(17)
  2	≥	2
  1	>	1

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