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

         ↳Narrowing Transformation

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))

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

A_{_2ND}(cons(X, X1)) -> A_{_2ND}(cons1(mark(X), mark(X1)))

10 new Dependency Pairs are created:

A_{_2ND}(cons(2nd(X''), X1)) -> A_{_2ND}(cons1(a_{_2nd}(mark(X'')), mark(X1)))
A_{_2ND}(cons(from(X''), X1)) -> A_{_2ND}(cons1(a_{_from}(mark(X'')), mark(X1)))
A_{_2ND}(cons(cons(X1'', X2'), X1)) -> A_{_2ND}(cons1(cons(mark(X1''), X2'), mark(X1)))
A_{_2ND}(cons(s(X''), X1)) -> A_{_2ND}(cons1(s(mark(X'')), mark(X1)))
A_{_2ND}(cons(cons1(X1'', X2'), X1)) -> A_{_2ND}(cons1(cons1(mark(X1''), mark(X2')), mark(X1)))
A_{_2ND}(cons(X, 2nd(X''))) -> A_{_2ND}(cons1(mark(X), a_{_2nd}(mark(X''))))
A_{_2ND}(cons(X, from(X''))) -> A_{_2ND}(cons1(mark(X), a_{_from}(mark(X''))))
A_{_2ND}(cons(X, cons(X1'', X2'))) -> A_{_2ND}(cons1(mark(X), cons(mark(X1''), X2')))
A_{_2ND}(cons(X, s(X''))) -> A_{_2ND}(cons1(mark(X), s(mark(X''))))
A_{_2ND}(cons(X, cons1(X1'', X2'))) -> A_{_2ND}(cons1(mark(X), cons1(mark(X1''), mark(X2'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

A_{_2ND}(cons(X, cons(X1'', X2'))) -> A_{_2ND}(cons1(mark(X), cons(mark(X1''), X2')))
A_{_2ND}(cons(X, from(X''))) -> A_{_2ND}(cons1(mark(X), a_{_from}(mark(X''))))
A_{_2ND}(cons(X, 2nd(X''))) -> A_{_2ND}(cons1(mark(X), a_{_2nd}(mark(X''))))
A_{_2ND}(cons(cons1(X1'', X2'), X1)) -> A_{_2ND}(cons1(cons1(mark(X1''), mark(X2')), mark(X1)))
A_{_2ND}(cons(s(X''), X1)) -> A_{_2ND}(cons1(s(mark(X'')), mark(X1)))
A_{_2ND}(cons(cons(X1'', X2'), X1)) -> A_{_2ND}(cons1(cons(mark(X1''), X2'), mark(X1)))
A_{_2ND}(cons(from(X''), X1)) -> A_{_2ND}(cons1(a_{_from}(mark(X'')), mark(X1)))
A_{_2ND}(cons(2nd(X''), X1)) -> A_{_2ND}(cons1(a_{_2nd}(mark(X'')), mark(X1)))
A_{_2ND}(cons(X, X1)) -> MARK(X)
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}(cons1(X, cons(Y, Z))) -> MARK(Y)
MARK(2nd(X)) -> A_{_2ND}(mark(X))
A_{_2ND}(cons(X, X1)) -> MARK(X1)

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))

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

MARK(2nd(X)) -> A_{_2ND}(mark(X))

five new Dependency Pairs are created:

MARK(2nd(2nd(X''))) -> A_{_2ND}(a_{_2nd}(mark(X'')))
MARK(2nd(from(X''))) -> A_{_2ND}(a_{_from}(mark(X'')))
MARK(2nd(cons(X1', X2'))) -> A_{_2ND}(cons(mark(X1'), X2'))
MARK(2nd(s(X''))) -> A_{_2ND}(s(mark(X'')))
MARK(2nd(cons1(X1', X2'))) -> A_{_2ND}(cons1(mark(X1'), mark(X2')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(2nd(cons1(X1', X2'))) -> A_{_2ND}(cons1(mark(X1'), mark(X2')))
A_{_2ND}(cons(X, from(X''))) -> A_{_2ND}(cons1(mark(X), a_{_from}(mark(X''))))
A_{_2ND}(cons(X, 2nd(X''))) -> A_{_2ND}(cons1(mark(X), a_{_2nd}(mark(X''))))
A_{_2ND}(cons(cons1(X1'', X2'), X1)) -> A_{_2ND}(cons1(cons1(mark(X1''), mark(X2')), mark(X1)))
A_{_2ND}(cons(s(X''), X1)) -> A_{_2ND}(cons1(s(mark(X'')), mark(X1)))
A_{_2ND}(cons(cons(X1'', X2'), X1)) -> A_{_2ND}(cons1(cons(mark(X1''), X2'), mark(X1)))
A_{_2ND}(cons(from(X''), X1)) -> A_{_2ND}(cons1(a_{_from}(mark(X'')), mark(X1)))
A_{_2ND}(cons(2nd(X''), X1)) -> A_{_2ND}(cons1(a_{_2nd}(mark(X'')), mark(X1)))
MARK(2nd(cons(X1', X2'))) -> A_{_2ND}(cons(mark(X1'), X2'))
A_{_2ND}(cons(X, X1)) -> MARK(X1)
MARK(2nd(from(X''))) -> A_{_2ND}(a_{_from}(mark(X'')))
A_{_2ND}(cons(X, X1)) -> MARK(X)
MARK(2nd(2nd(X''))) -> A_{_2ND}(a_{_2nd}(mark(X'')))
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}(cons1(X, cons(Y, Z))) -> MARK(Y)
A_{_2ND}(cons(X, cons(X1'', X2'))) -> A_{_2ND}(cons1(mark(X), cons(mark(X1''), X2')))

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))

The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes