Term Rewriting System R:
[X, Y, Z]
f(X) -> cons(X, f(g(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, Z)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(X) -> F(g(X))
F(X) -> G(X)
G(s(X)) -> G(X)
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Remaining


Dependency Pair:

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


Rules:


f(X) -> cons(X, f(g(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, Z)


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 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


f(X) -> cons(X, f(g(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, Z)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Remaining


Dependency Pair:

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


Rules:


f(X) -> cons(X, f(g(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, Z)


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


f(X) -> cons(X, f(g(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, Z)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

F(X) -> F(g(X))


Rules:


f(X) -> cons(X, f(g(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, Z)


Strategy:

innermost



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