Termination Proof using AProVE

Term Rewriting System R:
[Z, X, Y, X1, X2]
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ADD(s(X), Y) -> S(n_{_add}(activate(X), Y))
ADD(s(X), Y) -> ACTIVATE(X)
LEN(cons(X, Z)) -> S(n_{_len}(activate(Z)))
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_fst}(X1, X2)) -> FST(activate(X1), activate(X2))
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_len}(X)) -> LEN(activate(X))
ACTIVATE(n_{_len}(X)) -> ACTIVATE(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Negative Polynomial Order

Dependency Pairs:

ACTIVATE(n_{_len}(X)) -> ACTIVATE(X)
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_len}(X)) -> LEN(activate(X))
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_fst}(X1, X2)) -> FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

The following Dependency Pairs can be strictly oriented using the given order.

ACTIVATE(n_{_len}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_len}(X)) -> LEN(activate(X))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
s(X) -> n_{_s}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)

Used ordering:
Polynomial Order with Interpretation:

POL( ACTIVATE(x₁) ) = x₁

POL( n_{_len}(x₁) ) = x₁ + 1

POL( n_{_fst}(x₁, x₂) ) = x₁ + x₂

POL( FST(x₁, x₂) ) = x₁ + x₂

POL( activate(x₁) ) = x₁

POL( s(x₁) ) = x₁

POL( LEN(x₁) ) = x₁

POL( n_{_add}(x₁, x₂) ) = x₁ + x₂

POL( ADD(x₁, x₂) ) = x₁

POL( n_{_from}(x₁) ) = x₁

POL( cons(x₁, x₂) ) = x₂

POL( fst(x₁, x₂) ) = x₁ + x₂

POL( from(x₁) ) = x₁

POL( n_{_s}(x₁) ) = x₁

POL( add(x₁, x₂) ) = x₁ + x₂

POL( len(x₁) ) = x₁ + 1

POL( 0 ) = 0

POL( nil ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_fst}(X1, X2)) -> FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

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

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 3

                 ↳Negative Polynomial Order

Dependency Pairs:

FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_fst}(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(n_{_fst}(X1, X2)) -> FST(activate(X1), activate(X2))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

The following Dependency Pairs can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
s(X) -> n_{_s}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)

Used ordering:
Polynomial Order with Interpretation:

POL( FST(x₁, x₂) ) = x₁ + x₂ + 1

POL( s(x₁) ) = x₁

POL( cons(x₁, x₂) ) = x₂

POL( ACTIVATE(x₁) ) = x₁

POL( n_{_add}(x₁, x₂) ) = x₁ + x₂

POL( ADD(x₁, x₂) ) = x₁

POL( activate(x₁) ) = x₁

POL( n_{_fst}(x₁, x₂) ) = x₁ + x₂ + 1

POL( n_{_from}(x₁) ) = x₁

POL( fst(x₁, x₂) ) = x₁ + x₂ + 1

POL( from(x₁) ) = x₁

POL( n_{_s}(x₁) ) = x₁

POL( add(x₁, x₂) ) = x₁ + x₂

POL( n_{_len}(x₁) ) = 0

POL( len(x₁) ) = 0

POL( 0 ) = 0

POL( nil ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Dependency Graph

Dependency Pairs:

ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_fst}(X1, X2)) -> FST(activate(X1), activate(X2))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

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

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 5

                 ↳Negative Polynomial Order

Dependency Pairs:

ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

The following Dependency Pairs can be strictly oriented using the given order.

ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X2)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
s(X) -> n_{_s}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)

Used ordering:
Polynomial Order with Interpretation:

POL( ACTIVATE(x₁) ) = x₁

POL( n_{_add}(x₁, x₂) ) = x₁ + x₂ + 1

POL( ADD(x₁, x₂) ) = x₁

POL( s(x₁) ) = x₁

POL( n_{_from}(x₁) ) = x₁

POL( activate(x₁) ) = x₁

POL( n_{_fst}(x₁, x₂) ) = 0

POL( fst(x₁, x₂) ) = 0

POL( from(x₁) ) = x₁

POL( n_{_s}(x₁) ) = x₁

POL( add(x₁, x₂) ) = x₁ + x₂ + 1

POL( n_{_len}(x₁) ) = 0

POL( len(x₁) ) = 0

POL( nil ) = 0

POL( cons(x₁, x₂) ) = 0

POL( 0 ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pairs:

ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

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

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 7

                 ↳Size-Change Principle

Dependency Pair:

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

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, n_{_fst}(activate(X), activate(Z)))
fst(X1, X2) -> n_{_fst}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
add(0, X) -> X
add(s(X), Y) -> s(n_{_add}(activate(X), Y))
add(X1, X2) -> n_{_add}(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(n_{_len}(activate(Z)))
len(X) -> n_{_len}(X)
s(X) -> n_{_s}(X)
activate(n_{_fst}(X1, X2)) -> fst(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(X)
activate(n_{_add}(X1, X2)) -> add(activate(X1), activate(X2))
activate(n_{_len}(X)) -> len(activate(X))
activate(X) -> X

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

n_{_from}(x₁) -> n_{_from}(x₁)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:02 minutes