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
                 ↳Argument Filtering and 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'))


The following usable rules for innermost w.r.t. to the AFS can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{ns, s} > {activate, 0} > n0

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
ns(x1) -> ns(x1)
activate(x1) -> activate(x1)
0 -> 0
s(x1) -> s(x1)


   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
                 ↳Argument Filtering and 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'))


The following usable rules for innermost w.r.t. to the AFS can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{ns, s} > activate
{n0, 0}

resulting in one new DP problem.
Used Argument Filtering System:
GEQ(x1, x2) -> GEQ(x1, x2)
ns(x1) -> ns(x1)
activate(x1) -> activate(x1)
0 -> 0
s(x1) -> s(x1)


   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