Term Rewriting System R:
[X, X1, X2]
af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AF(s(0)) -> AF(ap(s(0)))
AF(s(0)) -> AP(s(0))
MARK(f(X)) -> AF(mark(X))
MARK(f(X)) -> MARK(X)
MARK(p(X)) -> AP(mark(X))
MARK(p(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

AF(s(0)) -> AF(ap(s(0)))


Rules:


af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))





The following dependency pair can be strictly oriented:

AF(s(0)) -> AF(ap(s(0)))


Additionally, the following rules can be oriented:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__p(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(A__F(x1))=  x1  
  POL(s(x1))=  1  
  POL(mark(x1))=  x1  
  POL(f(x1))=  0  
  POL(p(x1))=  0  
  POL(a__f(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:


af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(p(X)) -> MARK(X)
MARK(f(X)) -> MARK(X)


Rules:


af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))





The following dependency pair can be strictly oriented:

MARK(s(X)) -> MARK(X)


Additionally, the following rules can be oriented:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__p(x1))=  x1  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(mark(x1))=  x1  
  POL(f(x1))=  x1  
  POL(p(x1))=  x1  
  POL(a__f(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(p(X)) -> MARK(X)
MARK(f(X)) -> MARK(X)


Rules:


af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))





The following dependency pairs can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(f(X)) -> MARK(X)


Additionally, the following rules can be oriented:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__p(x1))=  x1  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1  
  POL(s(x1))=  0  
  POL(mark(x1))=  x1  
  POL(f(x1))=  1 + x1  
  POL(p(x1))=  x1  
  POL(a__f(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:

MARK(p(X)) -> MARK(X)


Rules:


af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))





The following dependency pair can be strictly oriented:

MARK(p(X)) -> MARK(X)


Additionally, the following rules can be oriented:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__p(x1))=  1 + x1  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(cons(x1, x2))=  0  
  POL(s(x1))=  0  
  POL(mark(x1))=  x1  
  POL(f(x1))=  0  
  POL(p(x1))=  1 + x1  
  POL(a__f(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Polo
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pair:


Rules:


af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))





Using the Dependency Graph resulted in no new DP problems.

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