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
         ↳Polynomial Ordering
       →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





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__f(x1))=  0  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(n__s(x1))=  1  
  POL(n__0)=  0  
  POL(s(x1))=  1  
  POL(f(x1))=  0  
  POL(F(x1))=  x1  
  POL(p(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo


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.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →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(nf(X)) -> ACTIVATE(X)


Additionally, the following rules can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__f(x1))=  1 + x1  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(n__s(x1))=  x1  
  POL(n__0)=  0  
  POL(s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(f(x1))=  1 + x1  
  POL(p(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Polynomial Ordering


Dependency Pair:

ACTIVATE(ns(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)


Additionally, the following rules can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__f(x1))=  0  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(n__s(x1))=  1 + x1  
  POL(n__0)=  0  
  POL(s(x1))=  1 + x1  
  POL(ACTIVATE(x1))=  x1  
  POL(f(x1))=  0  
  POL(p(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Polo
             ...
               →DP Problem 5
                 ↳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