Term Rewriting System R:
[X]
f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(0)) -> F(p(s(0)))
F(s(0)) -> P(s(0))
ACTIVATE(nf(X)) -> F(activate(X))
ACTIVATE(nf(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(n0) -> 0'

Furthermore, R contains two SCCs. 


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


Dependency Pair:

F(s(0)) -> F(p(s(0)))


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





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

F(s(0)) -> F(p(s(0)))
three new Dependency Pairs are created:

F(s(0)) -> F(0)
F(s(0)) -> F(p(ns(0)))
F(s(0)) -> F(p(s(n0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 3
             ↳Narrowing Transformation
       →DP Problem 2
         ↳Polo


Dependency Pairs:

F(s(0)) -> F(p(s(n0)))
F(s(0)) -> F(p(ns(0)))
F(s(0)) -> F(0)


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





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

F(s(0)) -> F(0)
one new Dependency Pair is created:

F(s(0)) -> F(n0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 3
             ↳Nar
             ...
               →DP Problem 4
                 ↳Narrowing Transformation
       →DP Problem 2
         ↳Polo


Dependency Pairs:

F(s(0)) -> F(p(ns(0)))
F(s(0)) -> F(p(s(n0)))


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





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

F(s(0)) -> F(p(ns(0)))
one new Dependency Pair is created:

F(s(0)) -> F(p(ns(n0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 3
             ↳Nar
             ...
               →DP Problem 5
                 ↳Narrowing Transformation
       →DP Problem 2
         ↳Polo


Dependency Pairs:

F(s(0)) -> F(p(ns(n0)))
F(s(0)) -> F(p(s(n0)))


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





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

F(s(0)) -> F(p(s(n0)))
one new Dependency Pair is created:

F(s(0)) -> F(p(ns(n0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 3
             ↳Nar
             ...
               →DP Problem 6
                 ↳Narrowing Transformation
       →DP Problem 2
         ↳Polo


Dependency Pairs:

F(s(0)) -> F(p(ns(n0)))
F(s(0)) -> F(p(ns(n0)))


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





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

F(s(0)) -> F(p(ns(n0)))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



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


Dependency Pair:

F(s(0)) -> F(p(ns(n0)))


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





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

F(s(0)) -> F(p(ns(n0)))
no new Dependency Pairs are created.
The transformation is resulting in no new DP problems. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nf(X)) -> ACTIVATE(X)


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





The following dependency pair can be strictly oriented:

ACTIVATE(ns(X)) -> ACTIVATE(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__f(x1))=  x1  
  POL(n__s(x1))=  1 + x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Polo
           →DP Problem 8
             ↳Polynomial Ordering


Dependency Pair:

ACTIVATE(nf(X)) -> ACTIVATE(X)


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





The following dependency pair can be strictly oriented:

ACTIVATE(nf(X)) -> ACTIVATE(X)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__f(x1))=  1 + x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Polo
           →DP Problem 8
             ↳Polo
             ...
               →DP Problem 9
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(0) -> cons(0, nf(ns(n0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
s(X) -> ns(X)
0 -> n0
activate(nf(X)) -> f(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes