Term Rewriting System R:
[X, XS, N, X1, X2]
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(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.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

AFROM(X) -> MARK(X)
AAFTER(0, XS) -> MARK(XS)
AAFTER(s(N), cons(X, XS)) -> AAFTER(mark(N), mark(XS))
AAFTER(s(N), cons(X, XS)) -> MARK(N)
AAFTER(s(N), cons(X, XS)) -> MARK(XS)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(after(X1, X2)) -> AAFTER(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.


   R
DPs
       →DP Problem 1
Narrowing Transformation


Dependency Pairs:

AAFTER(s(N), cons(X, XS)) -> MARK(XS)
AAFTER(s(N), cons(X, XS)) -> MARK(N)
AAFTER(s(N), cons(X, XS)) -> AAFTER(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)
AAFTER(0, XS) -> MARK(XS)
MARK(after(X1, X2)) -> AAFTER(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AFROM(X) -> MARK(X)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





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

AAFTER(s(N), cons(X, XS)) -> AAFTER(mark(N), mark(XS))
10 new Dependency Pairs are created:

AAFTER(s(from(X'')), cons(X, XS)) -> AAFTER(afrom(mark(X'')), mark(XS))
AAFTER(s(after(X1', X2')), cons(X, XS)) -> AAFTER(aafter(mark(X1'), mark(X2')), mark(XS))
AAFTER(s(cons(X1', X2')), cons(X, XS)) -> AAFTER(cons(mark(X1'), X2'), mark(XS))
AAFTER(s(s(X'')), cons(X, XS)) -> AAFTER(s(mark(X'')), mark(XS))
AAFTER(s(0), cons(X, XS)) -> AAFTER(0, mark(XS))
AAFTER(s(N), cons(X, from(X''))) -> AAFTER(mark(N), afrom(mark(X'')))
AAFTER(s(N), cons(X, after(X1', X2'))) -> AAFTER(mark(N), aafter(mark(X1'), mark(X2')))
AAFTER(s(N), cons(X, cons(X1', X2'))) -> AAFTER(mark(N), cons(mark(X1'), X2'))
AAFTER(s(N), cons(X, s(X''))) -> AAFTER(mark(N), s(mark(X'')))
AAFTER(s(N), cons(X, 0)) -> AAFTER(mark(N), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Narrowing Transformation


Dependency Pairs:

AAFTER(s(N), cons(X, 0)) -> AAFTER(mark(N), 0)
AAFTER(s(N), cons(X, s(X''))) -> AAFTER(mark(N), s(mark(X'')))
AAFTER(s(N), cons(X, cons(X1', X2'))) -> AAFTER(mark(N), cons(mark(X1'), X2'))
AAFTER(s(N), cons(X, after(X1', X2'))) -> AAFTER(mark(N), aafter(mark(X1'), mark(X2')))
AAFTER(s(N), cons(X, from(X''))) -> AAFTER(mark(N), afrom(mark(X'')))
AAFTER(s(0), cons(X, XS)) -> AAFTER(0, mark(XS))
AAFTER(s(s(X'')), cons(X, XS)) -> AAFTER(s(mark(X'')), mark(XS))
AAFTER(s(after(X1', X2')), cons(X, XS)) -> AAFTER(aafter(mark(X1'), mark(X2')), mark(XS))
AAFTER(s(from(X'')), cons(X, XS)) -> AAFTER(afrom(mark(X'')), mark(XS))
AAFTER(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)
AAFTER(0, XS) -> MARK(XS)
MARK(after(X1, X2)) -> AAFTER(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AAFTER(s(N), cons(X, XS)) -> MARK(XS)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





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

MARK(after(X1, X2)) -> AAFTER(mark(X1), mark(X2))
10 new Dependency Pairs are created:

MARK(after(from(X'), X2)) -> AAFTER(afrom(mark(X')), mark(X2))
MARK(after(after(X1'', X2''), X2)) -> AAFTER(aafter(mark(X1''), mark(X2'')), mark(X2))
MARK(after(cons(X1'', X2''), X2)) -> AAFTER(cons(mark(X1''), X2''), mark(X2))
MARK(after(s(X'), X2)) -> AAFTER(s(mark(X')), mark(X2))
MARK(after(0, X2)) -> AAFTER(0, mark(X2))
MARK(after(X1, from(X'))) -> AAFTER(mark(X1), afrom(mark(X')))
MARK(after(X1, after(X1'', X2''))) -> AAFTER(mark(X1), aafter(mark(X1''), mark(X2'')))
MARK(after(X1, cons(X1'', X2''))) -> AAFTER(mark(X1), cons(mark(X1''), X2''))
MARK(after(X1, s(X'))) -> AAFTER(mark(X1), s(mark(X')))
MARK(after(X1, 0)) -> AAFTER(mark(X1), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 3
Negative Polynomial Order


Dependency Pairs:

MARK(after(X1, 0)) -> AAFTER(mark(X1), 0)
MARK(after(X1, s(X'))) -> AAFTER(mark(X1), s(mark(X')))
MARK(after(X1, cons(X1'', X2''))) -> AAFTER(mark(X1), cons(mark(X1''), X2''))
MARK(after(X1, after(X1'', X2''))) -> AAFTER(mark(X1), aafter(mark(X1''), mark(X2'')))
MARK(after(X1, from(X'))) -> AAFTER(mark(X1), afrom(mark(X')))
MARK(after(0, X2)) -> AAFTER(0, mark(X2))
AAFTER(s(N), cons(X, s(X''))) -> AAFTER(mark(N), s(mark(X'')))
AAFTER(s(N), cons(X, cons(X1', X2'))) -> AAFTER(mark(N), cons(mark(X1'), X2'))
AAFTER(s(N), cons(X, after(X1', X2'))) -> AAFTER(mark(N), aafter(mark(X1'), mark(X2')))
AAFTER(s(N), cons(X, from(X''))) -> AAFTER(mark(N), afrom(mark(X'')))
AAFTER(s(0), cons(X, XS)) -> AAFTER(0, mark(XS))
AAFTER(s(s(X'')), cons(X, XS)) -> AAFTER(s(mark(X'')), mark(XS))
AAFTER(s(after(X1', X2')), cons(X, XS)) -> AAFTER(aafter(mark(X1'), mark(X2')), mark(XS))
AAFTER(s(from(X'')), cons(X, XS)) -> AAFTER(afrom(mark(X'')), mark(XS))
MARK(after(s(X'), X2)) -> AAFTER(s(mark(X')), mark(X2))
AAFTER(s(N), cons(X, XS)) -> MARK(XS)
MARK(after(after(X1'', X2''), X2)) -> AAFTER(aafter(mark(X1''), mark(X2'')), mark(X2))
AAFTER(s(N), cons(X, XS)) -> MARK(N)
MARK(after(from(X'), X2)) -> AAFTER(afrom(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)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AAFTER(0, XS) -> MARK(XS)
AAFTER(s(N), cons(X, 0)) -> AAFTER(mark(N), 0)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0





The following Dependency Pairs can be strictly oriented using the given order.

AAFTER(s(from(X'')), cons(X, XS)) -> AAFTER(afrom(mark(X'')), mark(XS))
MARK(after(from(X'), X2)) -> AAFTER(afrom(mark(X')), mark(X2))


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)


Used ordering:
Polynomial Order with Interpretation:

POL( AAFTER(x1, x2) ) = x1

POL( s(x1) ) = 1

POL( afrom(x1) ) = 0

POL( mark(x1) ) = 1

POL( MARK(x1) ) = 1

POL( AFROM(x1) ) = 1

POL( 0 ) = 1

POL( aafter(x1, x2) ) = 1

POL( cons(x1, x2) ) = 0

POL( after(x1, x2) ) = 0

POL( from(x1) ) = 0


This results in one new DP problem.


   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

MARK(after(X1, 0)) -> AAFTER(mark(X1), 0)
MARK(after(X1, s(X'))) -> AAFTER(mark(X1), s(mark(X')))
MARK(after(X1, cons(X1'', X2''))) -> AAFTER(mark(X1), cons(mark(X1''), X2''))
MARK(after(X1, after(X1'', X2''))) -> AAFTER(mark(X1), aafter(mark(X1''), mark(X2'')))
MARK(after(X1, from(X'))) -> AAFTER(mark(X1), afrom(mark(X')))
MARK(after(0, X2)) -> AAFTER(0, mark(X2))
AAFTER(s(N), cons(X, s(X''))) -> AAFTER(mark(N), s(mark(X'')))
AAFTER(s(N), cons(X, cons(X1', X2'))) -> AAFTER(mark(N), cons(mark(X1'), X2'))
AAFTER(s(N), cons(X, after(X1', X2'))) -> AAFTER(mark(N), aafter(mark(X1'), mark(X2')))
AAFTER(s(N), cons(X, from(X''))) -> AAFTER(mark(N), afrom(mark(X'')))
AAFTER(s(0), cons(X, XS)) -> AAFTER(0, mark(XS))
AAFTER(s(s(X'')), cons(X, XS)) -> AAFTER(s(mark(X'')), mark(XS))
AAFTER(s(after(X1', X2')), cons(X, XS)) -> AAFTER(aafter(mark(X1'), mark(X2')), mark(XS))
MARK(after(s(X'), X2)) -> AAFTER(s(mark(X')), mark(X2))
AAFTER(s(N), cons(X, XS)) -> MARK(XS)
MARK(after(after(X1'', X2''), X2)) -> AAFTER(aafter(mark(X1''), mark(X2'')), mark(X2))
AAFTER(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)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AAFTER(0, XS) -> MARK(XS)
AAFTER(s(N), cons(X, 0)) -> AAFTER(mark(N), 0)


Rules:


afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
aafter(0, XS) -> mark(XS)
aafter(s(N), cons(X, XS)) -> aafter(mark(N), mark(XS))
aafter(X1, X2) -> after(X1, X2)
mark(from(X)) -> afrom(mark(X))
mark(after(X1, X2)) -> aafter(mark(X1), mark(X2))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0




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