Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, n_{_from}(s(X)))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))
LENGTH(n_{_cons}(X, Y)) -> ACTIVATE(Y)
LENGTH1(X) -> LENGTH(activate(X))
LENGTH1(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_nil}) -> NIL
ACTIVATE(n_{_cons}(X1, X2)) -> CONS(X1, X2)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
activate(n_{_nil}) -> nil
nil -> n_{_nil}

Strategy:

innermost

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

LENGTH1(X) -> LENGTH(activate(X))

four new Dependency Pairs are created:

LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(X1', X2'))
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(n_{_nil}) -> LENGTH(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH1(X'') -> LENGTH(X'')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(X1', X2'))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
activate(n_{_nil}) -> nil
nil -> n_{_nil}

Strategy:

innermost

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

LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(X1', X2'))

one new Dependency Pair is created:

LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))
LENGTH1(X'') -> LENGTH(X'')

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
activate(n_{_nil}) -> nil
nil -> n_{_nil}

Strategy:

innermost

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

LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

four new Dependency Pairs are created:

LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(X1', X2'))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(X1', X2'))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
activate(n_{_nil}) -> nil
nil -> n_{_nil}

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(X1', X2'))

one new Dependency Pair is created:

LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
activate(n_{_nil}) -> nil
nil -> n_{_nil}

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)

one new Dependency Pair is created:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Usable Rules (Innermost)

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
activate(n_{_nil}) -> nil
nil -> n_{_nil}

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH1(n_{_from}(X'')) -> LENGTH(from(X''))

two new Dependency Pairs are created:

LENGTH1(n_{_from}(X''')) -> LENGTH(cons(X''', n_{_from}(s(X'''))))
LENGTH1(n_{_from}(X''')) -> LENGTH(n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))
LENGTH1(n_{_from}(X''')) -> LENGTH(cons(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH1(n_{_from}(X''')) -> LENGTH(cons(X''', n_{_from}(s(X'''))))

one new Dependency Pair is created:

LENGTH1(n_{_from}(X''')) -> LENGTH(n_{_cons}(X''', n_{_from}(s(X'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH1(n_{_from}(X''')) -> LENGTH(n_{_cons}(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(X''))

two new Dependency Pairs are created:

LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(cons(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 11

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH1(n_{_from}(X''')) -> LENGTH(n_{_cons}(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_from}(X'''))
LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(cons(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(cons(X''', n_{_from}(s(X'''))))

one new Dependency Pair is created:

LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_cons}(X''', n_{_from}(s(X'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 12

                 ↳Usable Rules (Innermost)

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_cons}(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_from}(X'''))
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_from}(X''')) -> LENGTH(n_{_cons}(X''', n_{_from}(s(X'''))))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 13

                 ↳Non Termination

Dependency Pairs:

Rule:

none

Strategy:

innermost

Found an infinite P-chain over R:
P =

LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_cons}(X''', n_{_from}(s(X'''))))
LENGTH(n_{_cons}(X, n_{_from}(X'''))) -> LENGTH1(n_{_from}(X'''))
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(n_{_cons}(X1', X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(n_{_cons}(X1', X2'))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_from}(X''')) -> LENGTH(n_{_cons}(X''', n_{_from}(s(X'''))))

R = none

s = LENGTH1(n_{_cons}(X', n_{_from}(X'''')))
evaluates to t =LENGTH1(n_{_cons}(X'''', n_{_from}(s(X''''))))

Thus, s starts an infinite chain as s matches t.

Innermost Termination of R could not be shown.
Duration:
0:15 minutes