Termination Proof using AProVE

Term Rewriting System R:
[X, Y, L, X1, X2]
eq(n_₀, n_₀) -> true
eq(n_{_s}(X), n_{_s}(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, n_{_inf}(s(X)))
inf(X) -> n_{_inf}(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), n_{_take}(activate(X), activate(L)))
take(X1, X2) -> n_{_take}(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(n_{_length}(activate(L)))
length(X) -> n_{_length}(X)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(n_{_inf}(X)) -> inf(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(n_{_length}(X)) -> length(X)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

EQ(n_{_s}(X), n_{_s}(Y)) -> EQ(activate(X), activate(Y))
EQ(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(X)
EQ(n_{_s}(X), n_{_s}(Y)) -> ACTIVATE(Y)
INF(X) -> S(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)
LENGTH(nil) -> 0'
LENGTH(cons(X, L)) -> S(n_{_length}(activate(L)))
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(n_₀) -> 0'
ACTIVATE(n_{_s}(X)) -> S(X)
ACTIVATE(n_{_inf}(X)) -> INF(X)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)
ACTIVATE(n_{_length}(X)) -> LENGTH(X)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(n_{_length}(X)) -> LENGTH(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Rules:

eq(n_₀, n_₀) -> true
eq(n_{_s}(X), n_{_s}(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, n_{_inf}(s(X)))
inf(X) -> n_{_inf}(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), n_{_take}(activate(X), activate(L)))
take(X1, X2) -> n_{_take}(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(n_{_length}(activate(L)))
length(X) -> n_{_length}(X)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(n_{_inf}(X)) -> inf(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(n_{_length}(X)) -> length(X)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_length}(X)) -> LENGTH(X)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__length(x₁)) = 1 + x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(n__take(x₁, x₂)) = x₁ + x₂

POL(TAKE(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(LENGTH(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
LENGTH(cons(X, L)) -> ACTIVATE(L)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Rules:

eq(n_₀, n_₀) -> true
eq(n_{_s}(X), n_{_s}(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, n_{_inf}(s(X)))
inf(X) -> n_{_inf}(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), n_{_take}(activate(X), activate(L)))
take(X1, X2) -> n_{_take}(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(n_{_length}(activate(L)))
length(X) -> n_{_length}(X)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(n_{_inf}(X)) -> inf(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(n_{_length}(X)) -> length(X)
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

         ↳Remaining

Dependency Pairs:

TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)

Rules:

eq(n_₀, n_₀) -> true
eq(n_{_s}(X), n_{_s}(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, n_{_inf}(s(X)))
inf(X) -> n_{_inf}(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), n_{_take}(activate(X), activate(L)))
take(X1, X2) -> n_{_take}(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(n_{_length}(activate(L)))
length(X) -> n_{_length}(X)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(n_{_inf}(X)) -> inf(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(n_{_length}(X)) -> length(X)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = x₁ + x₂

POL(n__take(x₁, x₂)) = 1 + x₁ + x₂

POL(TAKE(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)

Rules:

eq(n_₀, n_₀) -> true
eq(n_{_s}(X), n_{_s}(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, n_{_inf}(s(X)))
inf(X) -> n_{_inf}(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), n_{_take}(activate(X), activate(L)))
take(X1, X2) -> n_{_take}(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(n_{_length}(activate(L)))
length(X) -> n_{_length}(X)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(n_{_inf}(X)) -> inf(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(n_{_length}(X)) -> length(X)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

EQ(n_{_s}(X), n_{_s}(Y)) -> EQ(activate(X), activate(Y))

Rules:

eq(n_₀, n_₀) -> true
eq(n_{_s}(X), n_{_s}(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, n_{_inf}(s(X)))
inf(X) -> n_{_inf}(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), n_{_take}(activate(X), activate(L)))
take(X1, X2) -> n_{_take}(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(n_{_length}(activate(L)))
length(X) -> n_{_length}(X)
0 -> n_₀
s(X) -> n_{_s}(X)
activate(n_₀) -> 0
activate(n_{_s}(X)) -> s(X)
activate(n_{_inf}(X)) -> inf(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(n_{_length}(X)) -> length(X)
activate(X) -> X

Termination of R could not be shown.
Duration:
0:00 minutes

POL(n__length(x₁))	= 1 + x₁
POL(cons(x₁, x₂))	= x₁ + x₂
POL(n__take(x₁, x₂))	= x₁ + x₂
POL(TAKE(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁
POL(LENGTH(x₁))	= x₁