Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
a_{_2nd}(cons1(X, cons(Y, Z))) -> mark(Y)
a_{_2nd}(cons(X, X1)) -> a_{_2nd}(cons1(mark(X), mark(X1)))
a_{_2nd}(X) -> 2nd(X)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
mark(2nd(X)) -> a_{_2nd}(mark(X))
mark(from(X)) -> a_{_from}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_2ND}(cons1(X, cons(Y, Z))) -> MARK(Y)
A_{_2ND}(cons(X, X1)) -> A_{_2ND}(cons1(mark(X), mark(X1)))
A_{_2ND}(cons(X, X1)) -> MARK(X)
A_{_2ND}(cons(X, X1)) -> MARK(X1)
A_{_FROM}(X) -> MARK(X)
MARK(2nd(X)) -> A_{_2ND}(mark(X))
MARK(2nd(X)) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(from(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
MARK(cons1(X1, X2)) -> MARK(X1)
MARK(cons1(X1, X2)) -> MARK(X2)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

A_{_2ND}(cons(X, X1)) -> MARK(X1)
MARK(cons1(X1, X2)) -> MARK(X2)
MARK(cons1(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(2nd(X)) -> MARK(X)
A_{_2ND}(cons(X, X1)) -> MARK(X)
A_{_2ND}(cons(X, X1)) -> A_{_2ND}(cons1(mark(X), mark(X1)))
MARK(2nd(X)) -> A_{_2ND}(mark(X))
A_{_2ND}(cons1(X, cons(Y, Z))) -> MARK(Y)

Rules:

a_{_2nd}(cons1(X, cons(Y, Z))) -> mark(Y)
a_{_2nd}(cons(X, X1)) -> a_{_2nd}(cons1(mark(X), mark(X1)))
a_{_2nd}(X) -> 2nd(X)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
mark(2nd(X)) -> a_{_2nd}(mark(X))
mark(from(X)) -> a_{_from}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))

Termination of R could not be shown.
Duration:
0:00 minutes