/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) TransformationProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) NonTerminationLoopProof [COMPLETE, 0 ms]
        (31) NO
    (32) QDP
        (33) UsableRulesProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) QReductionProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (38) YES


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))

The TRS R 2 is 
   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false

The signature Sigma is {eq_2, true, false}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))


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

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

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

   EQ(s(X), s(Y)) -> EQ(X, Y)
   INF(X) -> INF(s(X))
   TAKE(s(X), cons(Y, L)) -> TAKE(X, L)
   LENGTH(cons(X, L)) -> LENGTH(L)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(cons(X, L)) -> LENGTH(L)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(cons(X, L)) -> LENGTH(L)

R is empty.
The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(cons(X, L)) -> LENGTH(L)

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

(12) 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:
*LENGTH(cons(X, L)) -> LENGTH(L)
The graph contains the following edges 1 > 1


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

(13)
YES

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(s(X), cons(Y, L)) -> TAKE(X, L)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(s(X), cons(Y, L)) -> TAKE(X, L)

R is empty.
The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(s(X), cons(Y, L)) -> TAKE(X, L)

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

(19) 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:
*TAKE(s(X), cons(Y, L)) -> TAKE(X, L)
The graph contains the following edges 1 > 1, 2 > 2


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

(20)
YES

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INF(X) -> INF(s(X))

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(22) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INF(X) -> INF(s(X))

R is empty.
The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(24) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))


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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INF(X) -> INF(s(X))

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

(26) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule INF(X) -> INF(s(X)) we obtained the following new rules [LPAR04]:

   (INF(s(z0)) -> INF(s(s(z0))),INF(s(z0)) -> INF(s(s(z0))))


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INF(s(z0)) -> INF(s(s(z0)))

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

(28) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule INF(s(z0)) -> INF(s(s(z0))) we obtained the following new rules [LPAR04]:

   (INF(s(s(z0))) -> INF(s(s(s(z0)))),INF(s(s(z0))) -> INF(s(s(s(z0)))))


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

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INF(s(s(z0))) -> INF(s(s(s(z0))))

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

(30) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = INF(s(s(z0))) evaluates to  t =INF(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [z0 / s(z0)]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from INF(s(s(z0))) to INF(s(s(s(z0)))).




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

(31)
NO

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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(s(X), s(Y)) -> EQ(X, Y)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(s(X), s(Y)) -> eq(X, Y)
   eq(X, Y) -> false
   inf(X) -> cons(X, inf(s(X)))
   take(0, X) -> nil
   take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
   length(nil) -> 0
   length(cons(X, L)) -> s(length(L))

The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(33) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(s(X), s(Y)) -> EQ(X, Y)

R is empty.
The set Q consists of the following terms:

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))

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

(35) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   eq(x0, x1)
   inf(x0)
   take(0, x0)
   take(s(x0), cons(x1, x2))
   length(nil)
   length(cons(x0, x1))


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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(s(X), s(Y)) -> EQ(X, Y)

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

(37) 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:
*EQ(s(X), s(Y)) -> EQ(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


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

(38)
YES