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}(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)
activate(n_{_fst}(X1, X2)) -> fst(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_add}(X1, X2)) -> add(X1, X2)
activate(n_{_len}(X)) -> len(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) -> ACTIVATE(X)
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_fst}(X1, X2)) -> FST(X1, X2)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(X1, X2)
ACTIVATE(n_{_len}(X)) -> LEN(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

LEN(cons(X, Z)) -> ACTIVATE(Z)

Additionally, the following rules can be oriented:

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)
activate(n_{_fst}(X1, X2)) -> fst(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_add}(X1, X2)) -> add(X1, X2)
activate(n_{_len}(X)) -> len(X)
activate(X) -> X
from(X) -> cons(X, n_{_from}(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)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 0

POL(activate(x₁)) = x₁

POL(FST(x₁, x₂)) = x₁ + x₂

POL(len(x₁)) = 1 + x₁

POL(n__fst(x₁, x₂)) = x₁ + x₂

POL(LEN(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

POL(ADD(x₁, x₂)) = x₁

POL(add(x₁, x₂)) = x₁ + x₂

POL(n__from(x₁)) = 0

POL(0) = 0

POL(cons(x₁, x₂)) = x₂

POL(nil) = 0

POL(s(x₁)) = x₁

POL(fst(x₁, x₂)) = x₁ + x₂

POL(n__len(x₁)) = 1 + x₁

POL(n__add(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

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

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)
activate(n_{_fst}(X1, X2)) -> fst(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_add}(X1, X2)) -> add(X1, X2)
activate(n_{_len}(X)) -> len(X)
activate(X) -> X
from(X) -> cons(X, n_{_from}(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)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 0

POL(activate(x₁)) = x₁

POL(FST(x₁, x₂)) = 1 + x₁ + x₂

POL(len(x₁)) = 0

POL(n__fst(x₁, x₂)) = 1 + x₁ + x₂

POL(ACTIVATE(x₁)) = x₁

POL(ADD(x₁, x₂)) = x₁

POL(add(x₁, x₂)) = x₁ + x₂

POL(n__from(x₁)) = 0

POL(0) = 0

POL(cons(x₁, x₂)) = x₂

POL(nil) = 0

POL(s(x₁)) = x₁

POL(fst(x₁, x₂)) = 1 + x₁ + x₂

POL(n__len(x₁)) = 0

POL(n__add(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Dependency Graph

Dependency Pairs:

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

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(X1, 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}(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)
activate(n_{_fst}(X1, X2)) -> fst(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_add}(X1, X2)) -> add(X1, X2)
activate(n_{_len}(X)) -> len(X)
activate(X) -> X

The following dependency pair can be strictly oriented:

ADD(s(X), Y) -> ACTIVATE(X)

Additionally, the following rules can be oriented:

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)
activate(n_{_fst}(X1, X2)) -> fst(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_add}(X1, X2)) -> add(X1, X2)
activate(n_{_len}(X)) -> len(X)
activate(X) -> X
from(X) -> cons(X, n_{_from}(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)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 0

POL(activate(x₁)) = x₁

POL(len(x₁)) = 0

POL(n__fst(x₁, x₂)) = 0

POL(ACTIVATE(x₁)) = x₁

POL(ADD(x₁, x₂)) = 1 + x₁

POL(add(x₁, x₂)) = 1 + x₁ + x₂

POL(n__from(x₁)) = 0

POL(0) = 0

POL(cons(x₁, x₂)) = 0

POL(nil) = 0

POL(fst(x₁, x₂)) = 0

POL(s(x₁)) = x₁

POL(n__len(x₁)) = 0

POL(n__add(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

ACTIVATE(n_{_add}(X1, X2)) -> ADD(X1, 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}(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)
activate(n_{_fst}(X1, X2)) -> fst(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_add}(X1, X2)) -> add(X1, X2)
activate(n_{_len}(X)) -> len(X)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

POL(from(x₁))	= 0
POL(activate(x₁))	= x₁
POL(FST(x₁, x₂))	= x₁ + x₂
POL(len(x₁))	= 1 + x₁
POL(n__fst(x₁, x₂))	= x₁ + x₂
POL(LEN(x₁))	= 1 + x₁
POL(ACTIVATE(x₁))	= x₁
POL(ADD(x₁, x₂))	= x₁
POL(add(x₁, x₂))	= x₁ + x₂
POL(n__from(x₁))	= 0
POL(0)	= 0
POL(cons(x₁, x₂))	= x₂
POL(nil)	= 0
POL(s(x₁))	= x₁
POL(fst(x₁, x₂))	= x₁ + x₂
POL(n__len(x₁))	= 1 + x₁
POL(n__add(x₁, x₂))	= x₁ + x₂