Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N, X1, X2]
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_after}(0, XS) -> mark(XS)
a_{_after}(s(N), cons(X, XS)) -> a_{_after}(mark(N), mark(XS))
a_{_after}(X1, X2) -> after(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(after(X1, X2)) -> a_{_after}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_FROM}(X) -> MARK(X)
A_{_AFTER}(0, XS) -> MARK(XS)
A_{_AFTER}(s(N), cons(X, XS)) -> A_{_AFTER}(mark(N), mark(XS))
A_{_AFTER}(s(N), cons(X, XS)) -> MARK(N)
A_{_AFTER}(s(N), cons(X, XS)) -> MARK(XS)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(from(X)) -> MARK(X)
MARK(after(X1, X2)) -> A_{_AFTER}(mark(X1), mark(X2))
MARK(after(X1, X2)) -> MARK(X1)
MARK(after(X1, X2)) -> MARK(X2)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

A_{_AFTER}(s(N), cons(X, XS)) -> MARK(XS)
A_{_AFTER}(s(N), cons(X, XS)) -> MARK(N)
A_{_AFTER}(s(N), cons(X, XS)) -> A_{_AFTER}(mark(N), mark(XS))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(after(X1, X2)) -> MARK(X2)
MARK(after(X1, X2)) -> MARK(X1)
A_{_AFTER}(0, XS) -> MARK(XS)
MARK(after(X1, X2)) -> A_{_AFTER}(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
A_{_FROM}(X) -> MARK(X)

Rules:

a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_after}(0, XS) -> mark(XS)
a_{_after}(s(N), cons(X, XS)) -> a_{_after}(mark(N), mark(XS))
a_{_after}(X1, X2) -> after(X1, X2)
mark(from(X)) -> a_{_from}(mark(X))
mark(after(X1, X2)) -> a_{_after}(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Termination of R could not be shown.
Duration:
0:01 minutes