Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_FILTER}(cons(X, Y), s(N), M) -> MARK(X)
A_{_SIEVE}(cons(s(N), Y)) -> MARK(N)
A_{_NATS}(N) -> MARK(N)
A_{_ZPRIMES} -> A_{_SIEVE}(a_{_nats}(s(s(0))))
A_{_ZPRIMES} -> A_{_NATS}(s(s(0)))
MARK(filter(X1, X2, X3)) -> A_{_FILTER}(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(sieve(X)) -> A_{_SIEVE}(mark(X))
MARK(sieve(X)) -> MARK(X)
MARK(nats(X)) -> A_{_NATS}(mark(X))
MARK(nats(X)) -> MARK(X)
MARK(zprimes) -> A_{_ZPRIMES}
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Negative Polynomial Order

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
A_{_ZPRIMES} -> A_{_NATS}(s(s(0)))
A_{_ZPRIMES} -> A_{_SIEVE}(a_{_nats}(s(s(0))))
MARK(zprimes) -> A_{_ZPRIMES}
MARK(nats(X)) -> MARK(X)
A_{_NATS}(N) -> MARK(N)
MARK(nats(X)) -> A_{_NATS}(mark(X))
MARK(sieve(X)) -> MARK(X)
A_{_SIEVE}(cons(s(N), Y)) -> MARK(N)
MARK(sieve(X)) -> A_{_SIEVE}(mark(X))
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(filter(X1, X2, X3)) -> A_{_FILTER}(mark(X1), mark(X2), mark(X3))
A_{_FILTER}(cons(X, Y), s(N), M) -> MARK(X)

Rules:

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

Strategy:

innermost

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

MARK(cons(X1, X2)) -> MARK(X1)
A_{_ZPRIMES} -> A_{_NATS}(s(s(0)))
MARK(nats(X)) -> MARK(X)
MARK(nats(X)) -> A_{_NATS}(mark(X))
A_{_SIEVE}(cons(s(N), Y)) -> MARK(N)
A_{_FILTER}(cons(X, Y), s(N), M) -> MARK(X)

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = x₁

POL( cons(x₁, x₂) ) = x₁ + 1

POL( filter(x₁, ..., x₃) ) = x₁ + x₂ + x₃

POL( A_{_FILTER}(x₁, ..., x₃) ) = x₁

POL( mark(x₁) ) = x₁

POL( nats(x₁) ) = x₁ + 1

POL( A_{_NATS}(x₁) ) = x₁

POL( A_{_SIEVE}(x₁) ) = x₁

POL( s(x₁) ) = x₁

POL( sieve(x₁) ) = x₁

POL( A_{_ZPRIMES} ) = 1

POL( 0 ) = 0

POL( zprimes ) = 1

POL( a_{_nats}(x₁) ) = x₁ + 1

POL( a_{_sieve}(x₁) ) = x₁

POL( a_{_filter}(x₁, ..., x₃) ) = x₁ + x₂ + x₃

POL( a_{_zprimes} ) = 1

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

MARK(s(X)) -> MARK(X)
A_{_ZPRIMES} -> A_{_SIEVE}(a_{_nats}(s(s(0))))
MARK(zprimes) -> A_{_ZPRIMES}
A_{_NATS}(N) -> MARK(N)
MARK(sieve(X)) -> MARK(X)
MARK(sieve(X)) -> A_{_SIEVE}(mark(X))
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(filter(X1, X2, X3)) -> A_{_FILTER}(mark(X1), mark(X2), mark(X3))

Rules:

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

Strategy:

innermost

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

                 ↳Usable Rules (Innermost)

Dependency Pairs:

MARK(sieve(X)) -> MARK(X)
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 17 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳Neg POLO

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Size-Change Principle

Dependency Pairs:

MARK(sieve(X)) -> MARK(X)
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

MARK(sieve(X)) -> MARK(X)
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

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

{1, 2, 3, 4, 5}	,	{1, 2, 3, 4, 5}
1	>	1

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

{1, 2, 3, 4, 5}	,	{1, 2, 3, 4, 5}
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:

filter(x₁, x₂, x₃) -> filter(x₁, x₂, x₃)
sieve(x₁) -> sieve(x₁)
s(x₁) -> s(x₁)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:05 minutes