Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N, X1, X2]
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

2ND(cons(X, XS)) -> HEAD(activate(XS))
2ND(cons(X, XS)) -> ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(activate(X1), activate(X2))
ACTIVATE(n_{_take}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_take}(X1, X2)) -> ACTIVATE(X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pairs:

ACTIVATE(n_{_take}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_take}(X1, X2)) -> ACTIVATE(X1)
TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

The following dependency pairs can be strictly oriented:

ACTIVATE(n_{_take}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_take}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(activate(X1), activate(X2))

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

activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
s(X) -> n_{_s}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = x₁

POL(from(x₁)) = x₁

POL(activate(x₁)) = x₁

POL(0) = 0

POL(n__take(x₁, x₂)) = 1 + x₁ + x₂

POL(cons(x₁, x₂)) = x₂

POL(take(x₁, x₂)) = 1 + x₁ + x₂

POL(TAKE(x₁, x₂)) = x₂

POL(n__s(x₁)) = x₁

POL(nil) = 0

POL(s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

Dependency Pairs:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pairs:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)

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(n__from(x₁)) = x₁

POL(n__s(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pair:

ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

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(n__from(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

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(n__from(x₁)) = 0

POL(from(x₁)) = 0

POL(activate(x₁)) = 0

POL(0) = 0

POL(n__take(x₁, x₂)) = 0

POL(cons(x₁, x₂)) = 0

POL(SEL(x₁, x₂)) = x₁

POL(take(x₁, x₂)) = 0

POL(n__s(x₁)) = 0

POL(nil) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_take}(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

POL(n__from(x₁))	= x₁
POL(from(x₁))	= x₁
POL(activate(x₁))	= x₁
POL(0)	= 0
POL(n__take(x₁, x₂))	= 1 + x₁ + x₂
POL(cons(x₁, x₂))	= x₂
POL(take(x₁, x₂))	= 1 + x₁ + x₂
POL(TAKE(x₁, x₂))	= x₂
POL(n__s(x₁))	= x₁
POL(nil)	= 0
POL(s(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁

POL(n__from(x₁))	= 1 + x₁
POL(ACTIVATE(x₁))	= x₁