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))

Innermost 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
         ↳Narrowing Transformation
       →DP Problem 2
         ↳AFS


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))


Strategy:

innermost




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

AF(s(0)) -> AF(ap(s(0)))
two new Dependency Pairs are created:

AF(s(0)) -> AF(0)
AF(s(0)) -> AF(p(s(0)))

The transformation is resulting in no new DP problems. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Argument Filtering and 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))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS 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)
s(x1) -> s(x1)
cons(x1, x2) -> cons(x1, x2)
p(x1) -> p(x1)
f(x1) -> f(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳AFS
           →DP Problem 3
             ↳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))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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