Term Rewriting System R:
[X, Y, X1, X2]
af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AF(g(X), Y) -> AF(mark(X), f(g(X), Y))
AF(g(X), Y) -> MARK(X)
MARK(f(X1, X2)) -> AF(mark(X1), X2)
MARK(f(X1, X2)) -> MARK(X1)
MARK(g(X)) -> MARK(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

MARK(g(X)) -> MARK(X)
MARK(f(X1, X2)) -> MARK(X1)
MARK(f(X1, X2)) -> AF(mark(X1), X2)
AF(g(X), Y) -> MARK(X)
AF(g(X), Y) -> AF(mark(X), f(g(X), Y))


Rules:


af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))





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

MARK(f(X1, X2)) -> AF(mark(X1), X2)
two new Dependency Pairs are created:

MARK(f(f(X1'', X2''), X2)) -> AF(af(mark(X1''), X2''), X2)
MARK(f(g(X'), X2)) -> AF(g(mark(X')), X2)

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

MARK(f(g(X'), X2)) -> AF(g(mark(X')), X2)
AF(g(X), Y) -> MARK(X)
AF(g(X), Y) -> AF(mark(X), f(g(X), Y))
MARK(f(f(X1'', X2''), X2)) -> AF(af(mark(X1''), X2''), X2)
MARK(f(X1, X2)) -> MARK(X1)
MARK(g(X)) -> MARK(X)


Rules:


af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))





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

MARK(f(f(X1'', X2''), X2)) -> AF(af(mark(X1''), X2''), X2)
three new Dependency Pairs are created:

MARK(f(f(X1''', X2'''), X2)) -> AF(f(mark(X1'''), X2'''), X2)
MARK(f(f(f(X1', X2'''), X2''), X2)) -> AF(af(af(mark(X1'), X2'''), X2''), X2)
MARK(f(f(g(X'), X2''), X2)) -> AF(af(g(mark(X')), X2''), X2)

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

MARK(f(f(g(X'), X2''), X2)) -> AF(af(g(mark(X')), X2''), X2)
MARK(f(f(f(X1', X2'''), X2''), X2)) -> AF(af(af(mark(X1'), X2'''), X2''), X2)
MARK(g(X)) -> MARK(X)
MARK(f(X1, X2)) -> MARK(X1)
AF(g(X), Y) -> MARK(X)
AF(g(X), Y) -> AF(mark(X), f(g(X), Y))
MARK(f(g(X'), X2)) -> AF(g(mark(X')), X2)


Rules:


af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))





The following dependency pairs can be strictly oriented:

MARK(g(X)) -> MARK(X)
AF(g(X), Y) -> MARK(X)
AF(g(X), Y) -> AF(mark(X), f(g(X), Y))


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

mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))
af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(g(x1))=  1 + x1  
  POL(A__F(x1, x2))=  x1  
  POL(mark(x1))=  x1  
  POL(f(x1, x2))=  x1  
  POL(a__f(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pairs:

MARK(f(f(g(X'), X2''), X2)) -> AF(af(g(mark(X')), X2''), X2)
MARK(f(f(f(X1', X2'''), X2''), X2)) -> AF(af(af(mark(X1'), X2'''), X2''), X2)
MARK(f(X1, X2)) -> MARK(X1)
MARK(f(g(X'), X2)) -> AF(g(mark(X')), X2)


Rules:


af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))





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


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


Dependency Pair:

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


Rules:


af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))





The following dependency pair can be strictly oriented:

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


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(f(x1, x2))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


af(g(X), Y) -> af(mark(X), f(g(X), Y))
af(X1, X2) -> f(X1, X2)
mark(f(X1, X2)) -> af(mark(X1), X2)
mark(g(X)) -> g(mark(X))





Using the Dependency Graph resulted in no new DP problems.

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