Certification Problem

Input (TPDB TRS_Standard/AG01/#3.40)

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 3 components.

• The 1^st component contains the pair

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

  →

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

  (10)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the

  prec(quot#)	=	1		stat(quot#)	=	lex
  prec(s)	=	0		stat(s)	=	lex
  prec(0)	=	2		stat(0)	=	lex

  π(quot#)	=	[2,1]
  π(s)	=	[1]
  π(minus)	=	1
  π(0)	=	[]

  together with the usable rules

  minus(x,0)	→	x	(1)
  minus(s(x),s(y))	→	minus(x,y)	(2)

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

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

  →

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

  (10)

  could be deleted.

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

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

  →

  minus#(x,y)

  (9)

  1.1.2 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [s(x1)]	=	1 · x1
  [minus#(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.

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

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

• The 3^rd component contains the pair

  plus#(minus(x,s(0)),minus(y,s(s(z))))	→	plus#(minus(y,s(s(z))),minus(x,s(0)))	(13)
  plus#(s(x),y)	→	plus#(x,y)	(12)
  plus#(plus(x,s(0)),plus(y,s(s(z))))	→	plus#(plus(y,s(s(z))),plus(x,s(0)))	(14)

  1.1.3 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  together with the usable rules

  plus(0,y)	→	y	(5)
  plus(s(x),y)	→	s(plus(x,y))	(6)
  plus(plus(x,s(0)),plus(y,s(s(z))))	→	plus(plus(y,s(s(z))),plus(x,s(0)))	(8)
  plus(minus(x,s(0)),minus(y,s(s(z))))	→	plus(minus(y,s(s(z))),minus(x,s(0)))	(7)
  minus(s(x),s(y))	→	minus(x,y)	(2)
  minus(x,0)	→	x	(1)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.3.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the rule

  minus(x,0)

  →

  x

  (1)

  could be deleted.

  1.1.3.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the rule

  plus(0,y)

  →

  y

  (5)

  could be deleted.

  1.1.3.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pair

  plus#(s(x),y)

  →

  plus#(x,y)

  (12)

  and the rules

  plus(s(x),y)	→	s(plus(x,y))	(6)
  minus(s(x),s(y))	→	minus(x,y)	(2)

  could be deleted.

  1.1.3.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.