Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(n_{_from}(X)) -> FROM(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

Dependency Pairs:

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

one new Dependency Pair is created:

FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Rules:

first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

one new Dependency Pair is created:

ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))
FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))

Rules:

first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

FIRST(s(X), cons(Y, n_{_first}(X1'', X2''))) -> ACTIVATE(n_{_first}(X1'', X2''))

one new Dependency Pair is created:

FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))
ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))

Rules:

first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(n_{_first}(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))) -> FIRST(s(X''), cons(Y'', n_{_first}(X1'''', X2'''')))

one new Dependency Pair is created:

ACTIVATE(n_{_first}(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

ACTIVATE(n_{_first}(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))
FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))

Rules:

first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_first}(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))) -> FIRST(s(X'''), cons(Y''', n_{_first}(s(X''''''), cons(Y'''''', n_{_first}(X1'''''''', X2'''''''')))))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = x₂

POL(FIRST(x₁, x₂)) = x₂

POL(s(x₁)) = 0

POL(ACTIVATE(x₁)) = x₁

POL(n__first(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

FIRST(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_first}(s(X''''), cons(Y'''', n_{_first}(X1'''''', X2''''''))))

Rules:

first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
activate(n_{_first}(X1, X2)) -> first(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(cons(x₁, x₂))	= x₂
POL(FIRST(x₁, x₂))	= x₂
POL(s(x₁))	= 0
POL(ACTIVATE(x₁))	= x₁
POL(n__first(x₁, x₂))	= 1 + x₂