Term Rewriting System R:
[X, Y, X1, X2]
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AFROM(X) -> MARK(X)
ALENGTH(cons(X, Y)) -> ALENGTH1(Y)
ALENGTH1(X) -> ALENGTH(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(length(X)) -> ALENGTH(X)
MARK(length1(X)) -> ALENGTH1(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 Pairs:

ALENGTH1(X) -> ALENGTH(X)
ALENGTH(cons(X, Y)) -> ALENGTH1(Y)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0





The following dependency pair can be strictly oriented:

ALENGTH1(X) -> ALENGTH(X)


Additionally, the following rules can be oriented:

afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(a__length(x1))=  0  
  POL(mark(x1))=  1 + x1  
  POL(a__from(x1))=  1  
  POL(a__length1(x1))=  0  
  POL(length1(x1))=  0  
  POL(A__LENGTH1(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(A__LENGTH(x1))=  x1  
  POL(length(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:

ALENGTH(cons(X, Y)) -> ALENGTH1(Y)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0





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(from(X)) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AFROM(X) -> MARK(X)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0





The following dependency pairs can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))


Additionally, the following rules can be oriented:

afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(a__length(x1))=  0  
  POL(MARK(x1))=  x1  
  POL(A__FROM(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__from(x1))=  1 + x1  
  POL(a__length1(x1))=  0  
  POL(length1(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x1  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(length(x1))=  0  

resulting in one new DP problem. 



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


Dependency Pairs:

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


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0





Using the Dependency Graph the DP problem was split into 1 DP problems.


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


Dependency Pair:

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


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(a__length(x1))=  x1  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(cons(x1, x2))=  1 + x2  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(mark(x1))=  1 + x1  
  POL(a__from(x1))=  1  
  POL(a__length1(x1))=  x1  
  POL(length(x1))=  x1  
  POL(length1(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alength(nil) -> 0
alength(cons(X, Y)) -> s(alength1(Y))
alength(X) -> length(X)
alength1(X) -> alength(X)
alength1(X) -> length1(X)
mark(from(X)) -> afrom(mark(X))
mark(length(X)) -> alength(X)
mark(length1(X)) -> alength1(X)
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(nil) -> nil
mark(0) -> 0





Using the Dependency Graph resulted in no new DP problems.

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