Termination Proof

Input TRS

Termination of the rewrite relation of the following TRS is considered.

minus(x,0)	→	x
minus(s(x),s(y))	→	minus(x,y)
divide(x,y)	→	div(x,y,0)
div(0,s(y),z)	→	z
div(s(x),s(y),z)	→	div(minus(s(x),s(y)),s(y),s(z))

Proof

1 Dependency Pair Transformation

divide#(x,y)	→	div#(x,y,0)
div#(s(x),s(y),z)	→	div#(minus(s(x),s(y)),s(y),s(z))
div#(s(x),s(y),z)	→	minus#(s(x),s(y))
minus#(s(x),s(y))	→	minus#(x,y)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  div#(s(x),s(y),z)

  →

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

  1.1.1 Uncurrying Processor

  We uncurry the tuple-symbol div^# in combination with the following symbol map which also determines the applicative arities of these symbols.

  s(x1)	is mapped to	s(x1),	div-s#(x1, x2, x3)
  minus(x1, x2)	is mapped to	minus(x1, x2),	div-minus#(x1,...,x4)
  div(x1, x2, x3)	is mapped to	div(x1, x2, x3)
  divide(x1, x2)	is mapped to	divide(x1, x2)
  0	is mapped to	0

  This yields the following new rules in the eta-expansion

  div#(minus(x,0),z,u)	→	div#(x,z,u)
  div#(minus(s(x),s(y)),z,u)	→	div#(minus(x,y),z,u)

  Uncurrying the pairs and rules, and adding the uncurrying rules yields the pair(s)

  div-s#(x,s(y),z)	→	div-minus#(s(x),s(y),s(y),s(z))
  div#(s(x),y,z)	→	div-s#(x,y,z)
  div#(minus(x,y),z,u)	→	div-minus#(x,y,z,u)
  div-minus#(x,0,z,u)	→	div#(x,z,u)
  div-minus#(s(x),s(y),z,u)	→	div-minus#(x,y,z,u)

  and the set of rules

  minus(x,0)	→	x
  minus(s(x),s(y))	→	minus(x,y)
  divide(x,y)	→	div(x,y,0)
  div(0,s(y),z)	→	z
  div(s(x),s(y),z)	→	div(minus(s(x),s(y)),s(y),s(z))

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [minus(x1, x2)]	=	x1 + 1
  [div-s#(x1, x2, x3)]	=	x1 + 1
  [s(x1)]	=	x1 + 1
  [div-minus#(x1,...,x4)]	=	x1
  [div#(x1, x2, x3)]	=	x1

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

  div-s#(x,s(y),z)	→	div-minus#(s(x),s(y),s(y),s(z))
  div#(s(x),y,z)	→	div-s#(x,y,z)
  div-minus#(x,0,z,u)	→	div#(x,z,u)

  remain.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

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

  →

  minus#(x,y)

  1.1.2 Subterm Criterion Processor

  We use the projection π(minus^#) = 1 to remove all pairs.

  1.1.2.1 P is empty

  There are no pairs anymore.