Term Rewriting System R:
[Y, X, X1, X2]
minus(n0, Y) -> 0
minus(ns(X), ns(Y)) -> minus(activate(X), activate(Y))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
DIV(s(X), ns(Y)) -> GEQ(X, activate(Y))
DIV(s(X), ns(Y)) -> ACTIVATE(Y)
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(ndiv(X1, X2)) -> DIV(activate(X1), X2)
ACTIVATE(ndiv(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nminus(X1, X2)) -> MINUS(X1, X2)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation
       →DP Problem 2
         ↳Remaining


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))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
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))
10 new Dependency Pairs are created:

GEQ(ns(n0), ns(Y)) -> GEQ(0, activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(activate(X'')), activate(Y))
GEQ(ns(ndiv(X1', X2')), ns(Y)) -> GEQ(div(activate(X1'), X2'), activate(Y))
GEQ(ns(nminus(X1', X2')), ns(Y)) -> GEQ(minus(X1', X2'), 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(activate(X'')))
GEQ(ns(X), ns(ndiv(X1', X2'))) -> GEQ(activate(X), div(activate(X1'), X2'))
GEQ(ns(X), ns(nminus(X1', X2'))) -> GEQ(activate(X), minus(X1', X2'))
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 3
             ↳Rewriting Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

GEQ(ns(X), ns(Y')) -> GEQ(activate(X), Y')
GEQ(ns(X), ns(nminus(X1', X2'))) -> GEQ(activate(X), minus(X1', X2'))
GEQ(ns(X), ns(ndiv(X1', X2'))) -> GEQ(activate(X), div(activate(X1'), X2'))
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(activate(X'')))
GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(nminus(X1', X2')), ns(Y)) -> GEQ(minus(X1', X2'), activate(Y))
GEQ(ns(ndiv(X1', X2')), ns(Y)) -> GEQ(div(activate(X1'), X2'), activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(activate(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))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
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 3
             ↳Rw
             ...
               →DP Problem 4
                 ↳Rewriting Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

GEQ(ns(X), ns(nminus(X1', X2'))) -> GEQ(activate(X), minus(X1', X2'))
GEQ(ns(X), ns(ndiv(X1', X2'))) -> GEQ(activate(X), div(activate(X1'), X2'))
GEQ(ns(X), ns(ns(X''))) -> GEQ(activate(X), s(activate(X'')))
GEQ(ns(X), ns(n0)) -> GEQ(activate(X), 0)
GEQ(ns(X''), ns(Y)) -> GEQ(X'', activate(Y))
GEQ(ns(nminus(X1', X2')), ns(Y)) -> GEQ(minus(X1', X2'), activate(Y))
GEQ(ns(ndiv(X1', X2')), ns(Y)) -> GEQ(div(activate(X1'), X2'), activate(Y))
GEQ(ns(ns(X'')), ns(Y)) -> GEQ(s(activate(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))
minus(X1, X2) -> nminus(X1, X2)
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(ndiv(nminus(X, activate(Y)), ns(activate(Y)))), n0)
div(X1, X2) -> ndiv(X1, X2)
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(activate(X))
activate(ndiv(X1, X2)) -> div(activate(X1), X2)
activate(nminus(X1, X2)) -> minus(X1, X2)
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
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:18 minutes