Termination Proof using AProVE

Term Rewriting System R:
[X, Y, M, N, X1, X2, X3]
filter(cons(X, Y), 0, M) -> cons(0, n_{_filter}(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, n_{_filter}(activate(Y), N, M))
filter(X1, X2, X3) -> n_{_filter}(X1, X2, X3)
sieve(cons(0, Y)) -> cons(0, n_{_sieve}(activate(Y)))
sieve(cons(s(N), Y)) -> cons(s(N), n_{_sieve}(filter(activate(Y), N, N)))
sieve(X) -> n_{_sieve}(X)
nats(N) -> cons(N, n_{_nats}(s(N)))
nats(X) -> n_{_nats}(X)
zprimes -> sieve(nats(s(s(0))))
activate(n_{_filter}(X1, X2, X3)) -> filter(X1, X2, X3)
activate(n_{_sieve}(X)) -> sieve(X)
activate(n_{_nats}(X)) -> nats(X)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)
SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
ZPRIMES -> SIEVE(nats(s(s(0))))
ZPRIMES -> NATS(s(s(0)))
ACTIVATE(n_{_filter}(X1, X2, X3)) -> FILTER(X1, X2, X3)
ACTIVATE(n_{_sieve}(X)) -> SIEVE(X)
ACTIVATE(n_{_nats}(X)) -> NATS(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

Dependency Pairs:

SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)
ACTIVATE(n_{_sieve}(X)) -> SIEVE(X)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
ACTIVATE(n_{_filter}(X1, X2, X3)) -> FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)

Rules:

filter(cons(X, Y), 0, M) -> cons(0, n_{_filter}(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, n_{_filter}(activate(Y), N, M))
filter(X1, X2, X3) -> n_{_filter}(X1, X2, X3)
sieve(cons(0, Y)) -> cons(0, n_{_sieve}(activate(Y)))
sieve(cons(s(N), Y)) -> cons(s(N), n_{_sieve}(filter(activate(Y), N, N)))
sieve(X) -> n_{_sieve}(X)
nats(N) -> cons(N, n_{_nats}(s(N)))
nats(X) -> n_{_nats}(X)
zprimes -> sieve(nats(s(s(0))))
activate(n_{_filter}(X1, X2, X3)) -> filter(X1, X2, X3)
activate(n_{_sieve}(X)) -> sieve(X)
activate(n_{_nats}(X)) -> nats(X)
activate(X) -> X

Strategy:

innermost

As we are in the innermost case, we can delete all 1 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Negative Polynomial Order

Dependency Pairs:

Rules:

nats(N) -> cons(N, n_{_nats}(s(N)))
nats(X) -> n_{_nats}(X)
activate(n_{_sieve}(X)) -> sieve(X)
activate(n_{_nats}(X)) -> nats(X)
activate(X) -> X
activate(n_{_filter}(X1, X2, X3)) -> filter(X1, X2, X3)
sieve(cons(s(N), Y)) -> cons(s(N), n_{_sieve}(filter(activate(Y), N, N)))
sieve(X) -> n_{_sieve}(X)
sieve(cons(0, Y)) -> cons(0, n_{_sieve}(activate(Y)))
filter(cons(X, Y), 0, M) -> cons(0, n_{_filter}(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, n_{_filter}(activate(Y), N, M))
filter(X1, X2, X3) -> n_{_filter}(X1, X2, X3)

Strategy:

innermost

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

SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)

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

nats(N) -> cons(N, n_{_nats}(s(N)))
nats(X) -> n_{_nats}(X)
activate(n_{_sieve}(X)) -> sieve(X)
activate(n_{_nats}(X)) -> nats(X)
activate(X) -> X
activate(n_{_filter}(X1, X2, X3)) -> filter(X1, X2, X3)
sieve(cons(s(N), Y)) -> cons(s(N), n_{_sieve}(filter(activate(Y), N, N)))
sieve(X) -> n_{_sieve}(X)
sieve(cons(0, Y)) -> cons(0, n_{_sieve}(activate(Y)))
filter(cons(X, Y), 0, M) -> cons(0, n_{_filter}(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, n_{_filter}(activate(Y), N, M))
filter(X1, X2, X3) -> n_{_filter}(X1, X2, X3)

Used ordering:
Polynomial Order with Interpretation:

POL( SIEVE(x₁) ) = x₁ + 1

POL( cons(x₁, x₂) ) = x₂

POL( ACTIVATE(x₁) ) = x₁

POL( n_{_sieve}(x₁) ) = x₁ + 1

POL( FILTER(x₁, ..., x₃) ) = x₁

POL( n_{_filter}(x₁, ..., x₃) ) = x₁

POL( activate(x₁) ) = x₁

POL( nats(x₁) ) = 0

POL( n_{_nats}(x₁) ) = 0

POL( sieve(x₁) ) = x₁ + 1

POL( filter(x₁, ..., x₃) ) = x₁

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Neg POLO

...

               →DP Problem 3

                 ↳Dependency Graph

Dependency Pairs:

ACTIVATE(n_{_sieve}(X)) -> SIEVE(X)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
ACTIVATE(n_{_filter}(X1, X2, X3)) -> FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)

Rules:

nats(N) -> cons(N, n_{_nats}(s(N)))
nats(X) -> n_{_nats}(X)
activate(n_{_sieve}(X)) -> sieve(X)
activate(n_{_nats}(X)) -> nats(X)
activate(X) -> X
activate(n_{_filter}(X1, X2, X3)) -> filter(X1, X2, X3)
sieve(cons(s(N), Y)) -> cons(s(N), n_{_sieve}(filter(activate(Y), N, N)))
sieve(X) -> n_{_sieve}(X)
sieve(cons(0, Y)) -> cons(0, n_{_sieve}(activate(Y)))
filter(cons(X, Y), 0, M) -> cons(0, n_{_filter}(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, n_{_filter}(activate(Y), N, M))
filter(X1, X2, X3) -> n_{_filter}(X1, X2, X3)

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Neg POLO

...

               →DP Problem 4

                 ↳Usable Rules (Innermost)

Dependency Pairs:

FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
ACTIVATE(n_{_filter}(X1, X2, X3)) -> FILTER(X1, X2, X3)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)

Rules:

nats(N) -> cons(N, n_{_nats}(s(N)))
nats(X) -> n_{_nats}(X)
activate(n_{_sieve}(X)) -> sieve(X)
activate(n_{_nats}(X)) -> nats(X)
activate(X) -> X
activate(n_{_filter}(X1, X2, X3)) -> filter(X1, X2, X3)
sieve(cons(s(N), Y)) -> cons(s(N), n_{_sieve}(filter(activate(Y), N, N)))
sieve(X) -> n_{_sieve}(X)
sieve(cons(0, Y)) -> cons(0, n_{_sieve}(activate(Y)))
filter(cons(X, Y), 0, M) -> cons(0, n_{_filter}(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, n_{_filter}(activate(Y), N, M))
filter(X1, X2, X3) -> n_{_filter}(X1, X2, X3)

Strategy:

innermost

As we are in the innermost case, we can delete all 12 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 2

             ↳Neg POLO

...

               →DP Problem 5

                 ↳Size-Change Principle

Dependency Pairs:

FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
ACTIVATE(n_{_filter}(X1, X2, X3)) -> FILTER(X1, X2, X3)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
ACTIVATE(n_{_filter}(X1, X2, X3)) -> FILTER(X1, X2, X3)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)

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

{1, 3}	,	{1, 3}
1	>	1

{2}	,	{2}
1	>	1
1	>	2
1	>	3

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

{2}	,	{1, 3}
1	>	1

{1, 3}	,	{2}
1	>	1
1	>	2
1	>	3

D_P: empty set
Oriented Rules: none

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

trivial

with Argument Filtering System:

n_{_filter}(x₁, x₂, x₃) -> n_{_filter}(x₁, x₂, x₃)
cons(x₁, x₂) -> cons(x₁, x₂)
s(x₁) -> s(x₁)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes