Term Rewriting System R:
[X, Y, Z, X1, X2]
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
AFROM(X) -> MARK(X)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(first(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(A__FIRST(x1, x2))=  x2  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(A__FROM(x1))=  x1  
  POL(s(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__from(x1))=  1 + x1  
  POL(a__first(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



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


Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





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


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


Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(first(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(A__FIRST(x1, x2))=  x2  
  POL(cons(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(mark(x1))=  x1  
  POL(a__from(x1))=  x1  
  POL(a__first(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



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


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(cons(X1, X2)) -> MARK(X1)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





The following dependency pairs can be strictly oriented:

AFIRST(s(X), cons(Y, Z)) -> MARK(Y)
MARK(cons(X1, X2)) -> MARK(X1)


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(first(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(MARK(x1))=  x1  
  POL(A__FIRST(x1, x2))=  x2  
  POL(cons(x1, x2))=  1 + x1  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(mark(x1))=  x1  
  POL(a__from(x1))=  1 + x1  
  POL(a__first(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





The following dependency pairs can be strictly oriented:

MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(first(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(0) -> 0
mark(nil) -> nil
mark(s(X)) -> s(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)





Using the Dependency Graph resulted in no new DP problems.

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