Term Rewriting System R:
[X, Y, Z]
f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nf(X)) -> f(X)
activate(X) -> X

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(X) -> G(X)
G(s(X)) -> G(X)
SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nf(X)) -> F(X)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

G(s(X)) -> G(X)


Rules:


f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nf(X)) -> f(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

G(s(X)) -> G(X)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(G(x1))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nf(X)) -> f(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))


Rules:


f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nf(X)) -> f(X)
activate(X) -> X


Strategy:

innermost




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

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
two new Dependency Pairs are created:

SEL(s(X), cons(Y, nf(X''))) -> SEL(X, f(X''))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nf(X''))) -> SEL(X, f(X''))


Rules:


f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nf(X)) -> f(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')


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

f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__f(x1))=  0  
  POL(0)=  0  
  POL(g(x1))=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(SEL(x1, x2))=  1 + x2  
  POL(s(x1))=  0  
  POL(f(x1))=  1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Polo
             ...
               →DP Problem 5
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

SEL(s(X), cons(Y, nf(X''))) -> SEL(X, f(X''))


Rules:


f(X) -> cons(X, nf(g(X)))
f(X) -> nf(X)
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nf(X)) -> f(X)
activate(X) -> X


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes