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

Innermost 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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(n_{_length}(X)) -> LENGTH(X)

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(n_{_length}(X)) -> LENGTH(X)

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

ACTIVATE(x₁) -> x₁
n_{_length}(x₁) -> n_{_length}(x₁)
LENGTH(x₁) -> x₁
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

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

Strategy:

innermost

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

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

12 new Dependency Pairs are created:

EQ(n_{_s}(n_₀), n_{_s}(Y)) -> EQ(0, activate(Y))
EQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> EQ(s(X''), activate(Y))
EQ(n_{_s}(n_{_inf}(X'')), n_{_s}(Y)) -> EQ(inf(X''), activate(Y))
EQ(n_{_s}(n_{_take}(X1', X2')), n_{_s}(Y)) -> EQ(take(X1', X2'), activate(Y))
EQ(n_{_s}(n_{_length}(X'')), n_{_s}(Y)) -> EQ(length(X''), activate(Y))
EQ(n_{_s}(X''), n_{_s}(Y)) -> EQ(X'', activate(Y))
EQ(n_{_s}(X), n_{_s}(n_₀)) -> EQ(activate(X), 0)
EQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> EQ(activate(X), s(X''))
EQ(n_{_s}(X), n_{_s}(n_{_inf}(X''))) -> EQ(activate(X), inf(X''))
EQ(n_{_s}(X), n_{_s}(n_{_take}(X1', X2'))) -> EQ(activate(X), take(X1', X2'))
EQ(n_{_s}(X), n_{_s}(n_{_length}(X''))) -> EQ(activate(X), length(X''))
EQ(n_{_s}(X), n_{_s}(Y')) -> EQ(activate(X), Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rewriting Transformation

Dependency Pairs:

EQ(n_{_s}(X), n_{_s}(Y')) -> EQ(activate(X), Y')
EQ(n_{_s}(X), n_{_s}(n_{_length}(X''))) -> EQ(activate(X), length(X''))
EQ(n_{_s}(X), n_{_s}(n_{_take}(X1', X2'))) -> EQ(activate(X), take(X1', X2'))
EQ(n_{_s}(X), n_{_s}(n_{_inf}(X''))) -> EQ(activate(X), inf(X''))
EQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> EQ(activate(X), s(X''))
EQ(n_{_s}(X), n_{_s}(n_₀)) -> EQ(activate(X), 0)
EQ(n_{_s}(X''), n_{_s}(Y)) -> EQ(X'', activate(Y))
EQ(n_{_s}(n_{_length}(X'')), n_{_s}(Y)) -> EQ(length(X''), activate(Y))
EQ(n_{_s}(n_{_take}(X1', X2')), n_{_s}(Y)) -> EQ(take(X1', X2'), activate(Y))
EQ(n_{_s}(n_{_inf}(X'')), n_{_s}(Y)) -> EQ(inf(X''), activate(Y))
EQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> EQ(s(X''), activate(Y))
EQ(n_{_s}(n_₀), n_{_s}(Y)) -> EQ(0, 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

Strategy:

innermost

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

EQ(n_{_s}(n_₀), n_{_s}(Y)) -> EQ(0, activate(Y))

one new Dependency Pair is created:

EQ(n_{_s}(n_₀), n_{_s}(Y)) -> EQ(n_₀, activate(Y))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 5

                 ↳Rewriting Transformation

Dependency Pairs:

EQ(n_{_s}(X), n_{_s}(n_{_length}(X''))) -> EQ(activate(X), length(X''))
EQ(n_{_s}(X), n_{_s}(n_{_take}(X1', X2'))) -> EQ(activate(X), take(X1', X2'))
EQ(n_{_s}(X), n_{_s}(n_{_inf}(X''))) -> EQ(activate(X), inf(X''))
EQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> EQ(activate(X), s(X''))
EQ(n_{_s}(X), n_{_s}(n_₀)) -> EQ(activate(X), 0)
EQ(n_{_s}(X''), n_{_s}(Y)) -> EQ(X'', activate(Y))
EQ(n_{_s}(n_{_length}(X'')), n_{_s}(Y)) -> EQ(length(X''), activate(Y))
EQ(n_{_s}(n_{_take}(X1', X2')), n_{_s}(Y)) -> EQ(take(X1', X2'), activate(Y))
EQ(n_{_s}(n_{_inf}(X'')), n_{_s}(Y)) -> EQ(inf(X''), activate(Y))
EQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> EQ(s(X''), activate(Y))
EQ(n_{_s}(X), n_{_s}(Y')) -> EQ(activate(X), 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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 6

                 ↳Rewriting Transformation

Dependency Pairs:

EQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> EQ(n_{_s}(X''), activate(Y))
EQ(n_{_s}(X), n_{_s}(Y')) -> EQ(activate(X), Y')
EQ(n_{_s}(X), n_{_s}(n_{_take}(X1', X2'))) -> EQ(activate(X), take(X1', X2'))
EQ(n_{_s}(X), n_{_s}(n_{_inf}(X''))) -> EQ(activate(X), inf(X''))
EQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> EQ(activate(X), s(X''))
EQ(n_{_s}(X), n_{_s}(n_₀)) -> EQ(activate(X), 0)
EQ(n_{_s}(X''), n_{_s}(Y)) -> EQ(X'', activate(Y))
EQ(n_{_s}(n_{_length}(X'')), n_{_s}(Y)) -> EQ(length(X''), activate(Y))
EQ(n_{_s}(n_{_take}(X1', X2')), n_{_s}(Y)) -> EQ(take(X1', X2'), activate(Y))
EQ(n_{_s}(n_{_inf}(X'')), n_{_s}(Y)) -> EQ(inf(X''), activate(Y))
EQ(n_{_s}(X), n_{_s}(n_{_length}(X''))) -> EQ(activate(X), length(X''))

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

Strategy:

innermost

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

EQ(n_{_s}(X), n_{_s}(n_₀)) -> EQ(activate(X), 0)

one new Dependency Pair is created:

EQ(n_{_s}(X), n_{_s}(n_₀)) -> EQ(activate(X), n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 7

                 ↳Rewriting Transformation

Dependency Pairs:

EQ(n_{_s}(X), n_{_s}(Y')) -> EQ(activate(X), Y')
EQ(n_{_s}(X), n_{_s}(n_{_length}(X''))) -> EQ(activate(X), length(X''))
EQ(n_{_s}(X), n_{_s}(n_{_take}(X1', X2'))) -> EQ(activate(X), take(X1', X2'))
EQ(n_{_s}(X), n_{_s}(n_{_inf}(X''))) -> EQ(activate(X), inf(X''))
EQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> EQ(activate(X), s(X''))
EQ(n_{_s}(X''), n_{_s}(Y)) -> EQ(X'', activate(Y))
EQ(n_{_s}(n_{_length}(X'')), n_{_s}(Y)) -> EQ(length(X''), activate(Y))
EQ(n_{_s}(n_{_take}(X1', X2')), n_{_s}(Y)) -> EQ(take(X1', X2'), activate(Y))
EQ(n_{_s}(n_{_inf}(X'')), n_{_s}(Y)) -> EQ(inf(X''), activate(Y))
EQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> EQ(n_{_s}(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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 8

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

EQ(n_{_s}(X), n_{_s}(n_{_s}(X''))) -> EQ(activate(X), n_{_s}(X''))
EQ(n_{_s}(n_{_s}(X'')), n_{_s}(Y)) -> EQ(n_{_s}(X''), activate(Y))
EQ(n_{_s}(X), n_{_s}(n_{_length}(X''))) -> EQ(activate(X), length(X''))
EQ(n_{_s}(X), n_{_s}(n_{_take}(X1', X2'))) -> EQ(activate(X), take(X1', X2'))
EQ(n_{_s}(X), n_{_s}(n_{_inf}(X''))) -> EQ(activate(X), inf(X''))
EQ(n_{_s}(X''), n_{_s}(Y)) -> EQ(X'', activate(Y))
EQ(n_{_s}(n_{_length}(X'')), n_{_s}(Y)) -> EQ(length(X''), activate(Y))
EQ(n_{_s}(n_{_take}(X1', X2')), n_{_s}(Y)) -> EQ(take(X1', X2'), activate(Y))
EQ(n_{_s}(n_{_inf}(X'')), n_{_s}(Y)) -> EQ(inf(X''), activate(Y))
EQ(n_{_s}(X), n_{_s}(Y')) -> EQ(activate(X), 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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:37 minutes