Term Rewriting System R:
[Z, X, Y, X1, X2]
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

Termination of R to be shown.

R
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)
LEN(cons(X, Z)) -> S(nlen(activate(Z)))
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(nlen(X)) -> LEN(activate(X))
ACTIVATE(nlen(X)) -> ACTIVATE(X)

Furthermore, R contains one SCC.

R
DPs
→DP Problem 1
Negative Polynomial Order

Dependency Pairs:

ACTIVATE(nlen(X)) -> ACTIVATE(X)
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nlen(X)) -> LEN(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(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, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

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

ACTIVATE(nlen(X)) -> ACTIVATE(X)
ACTIVATE(nlen(X)) -> LEN(activate(X))

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

activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
s(X) -> ns(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)

Used ordering:
Polynomial Order with Interpretation:

POL( ACTIVATE(x1) ) = x1

POL( nlen(x1) ) = x1 + 1

POL( nfst(x1, x2) ) = x1 + x2

POL( FST(x1, x2) ) = x1 + x2

POL( activate(x1) ) = x1

POL( s(x1) ) = x1

POL( LEN(x1) ) = x1

POL( nadd(x1, x2) ) = x1 + x2

POL( ADD(x1, x2) ) = x1

POL( nfrom(x1) ) = x1

POL( cons(x1, x2) ) = x2

POL( fst(x1, x2) ) = x1 + x2

POL( from(x1) ) = x1

POL( ns(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.

R
DPs
→DP Problem 1
Neg POLO
→DP Problem 2
Dependency Graph

Dependency Pairs:

LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(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, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

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

R
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(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(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(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)

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

activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
s(X) -> ns(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(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( nadd(x1, x2) ) = x1 + x2

POL( ADD(x1, x2) ) = x1

POL( activate(x1) ) = x1

POL( nfst(x1, x2) ) = x1 + x2 + 1

POL( nfrom(x1) ) = x1

POL( fst(x1, x2) ) = x1 + x2 + 1

POL( from(x1) ) = x1

POL( ns(x1) ) = x1

POL( add(x1, x2) ) = x1 + x2

POL( nlen(x1) ) = 0

POL( len(x1) ) = 0

POL( 0 ) = 0

POL( nil ) = 0

This results in one new DP problem.

R
DPs
→DP Problem 1
Neg POLO
→DP Problem 2
DGraph
...
→DP Problem 4
Dependency Graph

Dependency Pairs:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

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

R
DPs
→DP Problem 1
Neg POLO
→DP Problem 2
DGraph
...
→DP Problem 5
Negative Polynomial Order

Dependency Pairs:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

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

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

activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
s(X) -> ns(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)

Used ordering:
Polynomial Order with Interpretation:

POL( ACTIVATE(x1) ) = x1

POL( nadd(x1, x2) ) = x1 + x2 + 1

POL( ADD(x1, x2) ) = x1

POL( s(x1) ) = x1

POL( nfrom(x1) ) = x1

POL( activate(x1) ) = x1

POL( nfst(x1, x2) ) = 0

POL( fst(x1, x2) ) = 0

POL( from(x1) ) = x1

POL( ns(x1) ) = x1

POL( add(x1, x2) ) = x1 + x2 + 1

POL( nlen(x1) ) = 0

POL( len(x1) ) = 0

POL( nil ) = 0

POL( cons(x1, x2) ) = 0

POL( 0 ) = 0

This results in one new DP problem.

R
DPs
→DP Problem 1
Neg POLO
→DP Problem 2
DGraph
...
→DP Problem 6
Dependency Graph

Dependency Pairs:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

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

R
DPs
→DP Problem 1
Neg POLO
→DP Problem 2
DGraph
...
→DP Problem 7
Size-Change Principle

Dependency Pair:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

We number the DPs as follows:
1. ACTIVATE(nfrom(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

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
nfrom(x1) -> nfrom(x1)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:03 minutes