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))

Innermost 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
         ↳Argument Filtering and Ordering


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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(g(X)) -> MARK(X)


The following usable rules for innermost 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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{mark, MARK, g}

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
AF(x1, x2) -> x1
f(x1, x2) -> x1
mark(x1) -> mark(x1)
g(x1) -> g(x1)
af(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Argument Filtering and Ordering


Dependency Pairs:

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))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


The following usable rules for innermost 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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{MARK, g}

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
AF(x1, x2) -> x1
f(x1, x2) -> x1
mark(x1) -> x1
g(x1) -> g(x1)
af(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳AFS
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pairs:

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


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳AFS
             ...
               →DP Problem 4
                 ↳Argument Filtering and 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))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
f(x1, x2) -> f(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳AFS
             ...
               →DP Problem 5
                 ↳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))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes