YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 1 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 56 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following 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))
   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)))
   nats(N) -> cons(N, n__nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))
   filter(X1, X2, X3) -> n__filter(X1, X2, X3)
   sieve(X) -> n__sieve(X)
   nats(X) -> n__nats(X)
   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

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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)

The TRS R consists of the following 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))
   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)))
   nats(N) -> cons(N, n__nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))
   filter(X1, X2, X3) -> n__filter(X1, X2, X3)
   sieve(X) -> n__sieve(X)
   nats(X) -> n__nats(X)
   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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVATE(n__filter(X1, X2, X3)) -> FILTER(X1, X2, X3)
   FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
   ACTIVATE(n__sieve(X)) -> SIEVE(X)
   SIEVE(cons(0, Y)) -> ACTIVATE(Y)
   SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
   FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
   SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)

The TRS R consists of the following 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))
   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)))
   nats(N) -> cons(N, n__nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))
   filter(X1, X2, X3) -> n__filter(X1, X2, X3)
   sieve(X) -> n__sieve(X)
   nats(X) -> n__nats(X)
   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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   SIEVE(cons(0, Y)) -> ACTIVATE(Y)
   SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
   SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( FILTER_3(x_1, ..., x_3) ) = x_1
POL( activate_1(x_1) ) = x_1 + 2
POL( n__filter_3(x_1, ..., x_3) ) = max{0, x_1 - 2}
POL( filter_3(x_1, ..., x_3) ) = x_1
POL( n__sieve_1(x_1) ) = max{0, 2x_1 - 2}
POL( sieve_1(x_1) ) = 2x_1
POL( n__nats_1(x_1) ) = max{0, -2}
POL( nats_1(x_1) ) = 2
POL( cons_2(x_1, x_2) ) = x_2 + 2
POL( 0 ) = 0
POL( s_1(x_1) ) = 0
POL( ACTIVATE_1(x_1) ) = x_1 + 2
POL( SIEVE_1(x_1) ) = 2x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   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
   filter(X1, X2, X3) -> n__filter(X1, X2, X3)
   filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M))
   sieve(X) -> n__sieve(X)
   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)))
   filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M))
   nats(N) -> cons(N, n__nats(s(N)))
   nats(X) -> n__nats(X)


----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVATE(n__filter(X1, X2, X3)) -> FILTER(X1, X2, X3)
   FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
   ACTIVATE(n__sieve(X)) -> SIEVE(X)
   FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)

The TRS R consists of the following 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))
   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)))
   nats(N) -> cons(N, n__nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))
   filter(X1, X2, X3) -> n__filter(X1, X2, X3)
   sieve(X) -> n__sieve(X)
   nats(X) -> n__nats(X)
   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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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)

The TRS R consists of the following 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))
   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)))
   nats(N) -> cons(N, n__nats(s(N)))
   zprimes -> sieve(nats(s(s(0))))
   filter(X1, X2, X3) -> n__filter(X1, X2, X3)
   sieve(X) -> n__sieve(X)
   nats(X) -> n__nats(X)
   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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(9) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*ACTIVATE(n__filter(X1, X2, X3)) -> FILTER(X1, X2, X3)
The graph contains the following edges 1 > 1, 1 > 2, 1 > 3


*FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
The graph contains the following edges 1 > 1


*FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
The graph contains the following edges 1 > 1


----------------------------------------

(12)
YES