Termination Proof using AProVE

Term Rewriting System R:
[X, XS, X1, X2]
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(zeros) -> A_{_ZEROS}
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(tail(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
A_{_TAIL}(cons(X, XS)) -> MARK(XS)

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

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

MARK(tail(X)) -> A_{_TAIL}(mark(X))

four new Dependency Pairs are created:

MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(0)) -> A_{_TAIL}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})
MARK(tail(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

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

MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})

two new Dependency Pairs are created:

MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))
MARK(tail(zeros)) -> A_{_TAIL}(zeros)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

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

MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))

five new Dependency Pairs are created:

MARK(tail(tail(X'''))) -> A_{_TAIL}(tail(mark(X''')))
MARK(tail(tail(zeros))) -> A_{_TAIL}(a_{_tail}(a_{_zeros}))
MARK(tail(tail(tail(X')))) -> A_{_TAIL}(a_{_tail}(a_{_tail}(mark(X'))))
MARK(tail(tail(cons(X1', X2')))) -> A_{_TAIL}(a_{_tail}(cons(mark(X1'), X2')))
MARK(tail(tail(0))) -> A_{_TAIL}(a_{_tail}(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pairs:

MARK(tail(tail(0))) -> A_{_TAIL}(a_{_tail}(0))
MARK(tail(tail(cons(X1', X2')))) -> A_{_TAIL}(a_{_tail}(cons(mark(X1'), X2')))
MARK(tail(tail(tail(X')))) -> A_{_TAIL}(a_{_tail}(a_{_tail}(mark(X'))))
MARK(tail(tail(zeros))) -> A_{_TAIL}(a_{_tail}(a_{_zeros}))
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, XS)) -> MARK(XS)
MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

The following dependency pairs can be strictly oriented:

MARK(tail(tail(0))) -> A_{_TAIL}(a_{_tail}(0))
MARK(tail(tail(cons(X1', X2')))) -> A_{_TAIL}(a_{_tail}(cons(mark(X1'), X2')))
MARK(tail(tail(tail(X')))) -> A_{_TAIL}(a_{_tail}(a_{_tail}(mark(X'))))
MARK(tail(tail(zeros))) -> A_{_TAIL}(a_{_tail}(a_{_zeros}))
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(X)) -> MARK(X)
MARK(tail(zeros)) -> A_{_TAIL}(cons(0, zeros))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(MARK(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(a__zeros) = 0

POL(tail(x₁)) = 1 + x₁

POL(a__tail(x₁)) = 1 + x₁

POL(mark(x₁)) = x₁

POL(zeros) = 0

POL(A__TAIL(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Dependency Graph

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
A_{_TAIL}(cons(X, XS)) -> MARK(XS)

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Polynomial Ordering

Dependency Pair:

MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

The following dependency pair can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MARK(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pair:

Rules:

a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_tail}(cons(X, XS)) -> mark(XS)
a_{_tail}(X) -> tail(X)
mark(zeros) -> a_{_zeros}
mark(tail(X)) -> a_{_tail}(mark(X))
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(0)	= 0
POL(MARK(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁ + x₂
POL(a__zeros)	= 0
POL(tail(x₁))	= 1 + x₁
POL(a__tail(x₁))	= 1 + x₁
POL(mark(x₁))	= x₁
POL(zeros)	= 0
POL(A__TAIL(x₁))	= x₁