/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 12 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) NonTerminationLoopProof [COMPLETE, 0 ms]
        (38) NO


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

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

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

The set Q consists of the following terms:

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))


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

(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:

   FROM(X) -> FROM(s(X))
   SEL(s(N), cons(X, XS)) -> SEL(N, XS)
   MINUS(s(X), s(Y)) -> MINUS(X, Y)
   QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
   QUOT(s(X), s(Y)) -> MINUS(X, Y)
   ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y)
   ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

The set Q consists of the following terms:

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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 with 3 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

The set Q consists of the following terms:

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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:

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

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

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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].

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))


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

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

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

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


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

(13)
YES

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

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

   ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

The set Q consists of the following terms:

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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:

   ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

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

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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].

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))


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

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

   ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

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:
*ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)
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:

   SEL(s(N), cons(X, XS)) -> SEL(N, XS)

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

The set Q consists of the following terms:

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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:

   SEL(s(N), cons(X, XS)) -> SEL(N, XS)

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

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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].

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))


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

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

   SEL(s(N), cons(X, XS)) -> SEL(N, XS)

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

(26) 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:
*SEL(s(N), cons(X, XS)) -> SEL(N, XS)
The graph contains the following edges 1 > 1, 2 > 2


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

(27)
YES

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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(X) -> FROM(s(X))

The TRS R consists of the following rules:

   from(X) -> cons(X, from(s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, XS)
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

The set Q consists of the following terms:

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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

(29) 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.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(X) -> FROM(s(X))

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

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))

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

(31) 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].

   from(x0)
   sel(0, cons(x0, x1))
   sel(s(x0), cons(x1, x2))
   minus(x0, 0)
   minus(s(x0), s(x1))
   quot(0, s(x0))
   quot(s(x0), s(x1))
   zWquot(x0, nil)
   zWquot(nil, x0)
   zWquot(cons(x0, x1), cons(x2, x3))


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

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

   FROM(X) -> FROM(s(X))

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

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

   (FROM(s(z0)) -> FROM(s(s(z0))),FROM(s(z0)) -> FROM(s(s(z0))))


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

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

   FROM(s(z0)) -> FROM(s(s(z0)))

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

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

   (FROM(s(s(z0))) -> FROM(s(s(s(z0)))),FROM(s(s(z0))) -> FROM(s(s(s(z0)))))


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

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

   FROM(s(s(z0))) -> FROM(s(s(s(z0))))

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

(37) 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 = FROM(s(s(z0))) evaluates to  t =FROM(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 FROM(s(s(z0))) to FROM(s(s(s(z0)))).




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

(38)
NO