Term Rewriting System R:
[Y, X]
minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(n0, Y) -> 0'
MINUS(ns(X), ns(Y)) -> MINUS(activate(X), activate(Y))
MINUS(ns(X), ns(Y)) -> ACTIVATE(X)
MINUS(ns(X), ns(Y)) -> ACTIVATE(Y)
GEQ(ns(X), ns(Y)) -> GEQ(activate(X), activate(Y))
GEQ(ns(X), ns(Y)) -> ACTIVATE(X)
GEQ(ns(X), ns(Y)) -> ACTIVATE(Y)
DIV(s(X), ns(Y)) -> IF(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
DIV(s(X), ns(Y)) -> GEQ(X, activate(Y))
DIV(s(X), ns(Y)) -> ACTIVATE(Y)
DIV(s(X), ns(Y)) -> DIV(minus(X, activate(Y)), ns(activate(Y)))
DIV(s(X), ns(Y)) -> MINUS(X, activate(Y))
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(X)

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
Nar


Dependency Pair:

MINUS(ns(X), ns(Y)) -> MINUS(activate(X), activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X), ns(Y)) -> MINUS(activate(X), activate(Y))
six new Dependency Pairs are created:

MINUS(ns(n0), ns(Y)) -> MINUS(0, activate(Y))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(s(X''), activate(Y))
MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
MINUS(ns(X), ns(n0)) -> MINUS(activate(X), 0)
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), s(X''))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), s(X''))
MINUS(ns(X), ns(n0)) -> MINUS(activate(X), 0)
MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(s(X''), activate(Y))
MINUS(ns(n0), ns(Y)) -> MINUS(0, activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(n0), ns(Y)) -> MINUS(0, activate(Y))
one new Dependency Pair is created:

MINUS(ns(n0), ns(Y)) -> MINUS(n0, activate(Y))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 4
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), s(X''))
MINUS(ns(X), ns(n0)) -> MINUS(activate(X), 0)
MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(s(X''), activate(Y))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(ns(X'')), ns(Y)) -> MINUS(s(X''), activate(Y))
one new Dependency Pair is created:

MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 5
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
MINUS(ns(X), ns(n0)) -> MINUS(activate(X), 0)
MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), s(X''))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X), ns(n0)) -> MINUS(activate(X), 0)
one new Dependency Pair is created:

MINUS(ns(X), ns(n0)) -> MINUS(activate(X), n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 6
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), s(X''))
MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), s(X''))
one new Dependency Pair is created:

MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 7
Narrowing Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X''), ns(Y)) -> MINUS(X'', activate(Y))
three new Dependency Pairs are created:

MINUS(ns(X''), ns(n0)) -> MINUS(X'', 0)
MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', s(X'))
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 8
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', s(X'))
MINUS(ns(X''), ns(n0)) -> MINUS(X'', 0)
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X''), ns(n0)) -> MINUS(X'', 0)
one new Dependency Pair is created:

MINUS(ns(X''), ns(n0)) -> MINUS(X'', n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 9
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', s(X'))
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', s(X'))
one new Dependency Pair is created:

MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', ns(X'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 10
Narrowing Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', ns(X'))
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(X), ns(Y')) -> MINUS(activate(X), Y')
three new Dependency Pairs are created:

MINUS(ns(n0), ns(Y')) -> MINUS(0, Y')
MINUS(ns(ns(X'')), ns(Y')) -> MINUS(s(X''), Y')
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 11
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(ns(X'')), ns(Y')) -> MINUS(s(X''), Y')
MINUS(ns(n0), ns(Y')) -> MINUS(0, Y')
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', ns(X'))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(n0), ns(Y')) -> MINUS(0, Y')
one new Dependency Pair is created:

MINUS(ns(n0), ns(Y')) -> MINUS(n0, Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 12
Rewriting Transformation
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(ns(X'')), ns(Y')) -> MINUS(s(X''), Y')
MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', ns(X'))
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

MINUS(ns(ns(X'')), ns(Y')) -> MINUS(s(X''), Y')
one new Dependency Pair is created:

MINUS(ns(ns(X'')), ns(Y')) -> MINUS(ns(X''), Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 13
Polynomial Ordering
       →DP Problem 2
Nar


Dependency Pairs:

MINUS(ns(ns(X'')), ns(Y')) -> MINUS(ns(X''), Y')
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', ns(X'))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MINUS(ns(ns(X'')), ns(Y')) -> MINUS(ns(X''), Y')
MINUS(ns(X''), ns(Y')) -> MINUS(X'', Y')
MINUS(ns(X), ns(ns(X''))) -> MINUS(activate(X), ns(X''))
MINUS(ns(ns(X'')), ns(Y)) -> MINUS(ns(X''), activate(Y))
MINUS(ns(X''), ns(ns(X'))) -> MINUS(X'', ns(X'))


Additionally, the following usable rules for innermost can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X
0 -> n0
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(MINUS(x1, x2))=  x2  
  POL(n__s(x1))=  1 + x1  
  POL(n__0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 3
Rw
             ...
               →DP Problem 14
Dependency Graph
       →DP Problem 2
Nar


Dependency Pair:


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Narrowing Transformation


Dependency Pair:

GEQ(ns(X), ns(Y)) -> GEQ(activate(X), activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X), ns(Y)) -> GEQ(activate(X), activate(Y))
six new Dependency Pairs are created:

GEQ(ns(n0), ns(Y)) -> GEQ(0, activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(X''), activate(Y))
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(X''))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rewriting Transformation


Dependency Pairs:

GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(X''))
GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(X''), activate(Y))
GEQ(ns(n0), ns(Y)) -> GEQ(0, activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(n0), ns(Y)) -> GEQ(0, activate(Y))
one new Dependency Pair is created:

GEQ(ns(n0), ns(Y)) -> GEQ(n0, activate(Y))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 16
Rewriting Transformation


Dependency Pairs:

GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(X''))
GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(X''), activate(Y))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(X''), activate(Y))
one new Dependency Pair is created:

GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 17
Rewriting Transformation


Dependency Pairs:

GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(X''))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
one new Dependency Pair is created:

GEQ(ns(X), ns(n0)) -> GEQ(activate(X), n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 18
Rewriting Transformation


Dependency Pairs:

GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(X''))
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(X''))
one new Dependency Pair is created:

GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 19
Narrowing Transformation


Dependency Pairs:

GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
three new Dependency Pairs are created:

GEQ(ns(X''), ns(n0)) -> GEQ(X'', 0)
GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', s(X'))
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 20
Rewriting Transformation


Dependency Pairs:

GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', s(X'))
GEQ(ns(X''), ns(n0)) -> GEQ(X'', 0)
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X''), ns(n0)) -> GEQ(X'', 0)
one new Dependency Pair is created:

GEQ(ns(X''), ns(n0)) -> GEQ(X'', n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 21
Rewriting Transformation


Dependency Pairs:

GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', s(X'))
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', s(X'))
one new Dependency Pair is created:

GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', ns(X'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 22
Narrowing Transformation


Dependency Pairs:

GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', ns(X'))
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
three new Dependency Pairs are created:

GEQ(ns(n0), ns(Y')) -> GEQ(0, Y')
GEQ(ns(ns(X'')), ns(Y')) -> GEQ(s(X''), Y')
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 23
Rewriting Transformation


Dependency Pairs:

GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(ns(X'')), ns(Y')) -> GEQ(s(X''), Y')
GEQ(ns(n0), ns(Y')) -> GEQ(0, Y')
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', ns(X'))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(n0), ns(Y')) -> GEQ(0, Y')
one new Dependency Pair is created:

GEQ(ns(n0), ns(Y')) -> GEQ(n0, Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 24
Rewriting Transformation


Dependency Pairs:

GEQ(ns(ns(X'')), ns(Y')) -> GEQ(s(X''), Y')
GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', ns(X'))
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

GEQ(ns(ns(X'')), ns(Y')) -> GEQ(s(X''), Y')
one new Dependency Pair is created:

GEQ(ns(ns(X'')), ns(Y')) -> GEQ(ns(X''), Y')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 25
Polynomial Ordering


Dependency Pairs:

GEQ(ns(ns(X'')), ns(Y')) -> GEQ(ns(X''), Y')
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', ns(X'))


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

GEQ(ns(ns(X'')), ns(Y')) -> GEQ(ns(X''), Y')
GEQ(ns(X''), ns(Y')) -> GEQ(X'', Y')
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), ns(X''))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(ns(X''), activate(Y))
GEQ(ns(X''), ns(ns(X'))) -> GEQ(X'', ns(X'))


Additionally, the following usable rules for innermost can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X
0 -> n0
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(GEQ(x1, x2))=  x2  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(n__s(x1))=  1 + x1  
  POL(n__0)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 15
Rw
             ...
               →DP Problem 26
Dependency Graph


Dependency Pair:


Rules:


minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
geq(X, n0) -> true
geq(n0, ns(Y)) -> false
geq(ns(X), ns(Y)) -> geq(activate(X), activate(Y))
div(0, ns(Y)) -> 0
div(s(X), ns(Y)) -> if(geq(X, activate(Y)), ns(div(minus(X, activate(Y)), ns(activate(Y)))), n0)
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:05 minutes