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

Innermost 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

         ↳Narrowing Transformation

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

Strategy:

innermost

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

A_{_AFTER}(s(N), cons(X, XS)) -> A_{_AFTER}(mark(N), mark(XS))

10 new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

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

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

Strategy:

innermost

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

MARK(after(X1, X2)) -> A_{_AFTER}(mark(X1), mark(X2))

10 new Dependency Pairs are created:

MARK(after(from(X'), X2)) -> A_{_AFTER}(a_{_from}(mark(X')), mark(X2))
MARK(after(after(X1'', X2''), X2)) -> A_{_AFTER}(a_{_after}(mark(X1''), mark(X2'')), mark(X2))
MARK(after(cons(X1'', X2''), X2)) -> A_{_AFTER}(cons(mark(X1''), X2''), mark(X2))
MARK(after(s(X'), X2)) -> A_{_AFTER}(s(mark(X')), mark(X2))
MARK(after(0, X2)) -> A_{_AFTER}(0, mark(X2))
MARK(after(X1, from(X'))) -> A_{_AFTER}(mark(X1), a_{_from}(mark(X')))
MARK(after(X1, after(X1'', X2''))) -> A_{_AFTER}(mark(X1), a_{_after}(mark(X1''), mark(X2'')))
MARK(after(X1, cons(X1'', X2''))) -> A_{_AFTER}(mark(X1), cons(mark(X1''), X2''))
MARK(after(X1, s(X'))) -> A_{_AFTER}(mark(X1), s(mark(X')))
MARK(after(X1, 0)) -> A_{_AFTER}(mark(X1), 0)

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

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:28 minutes