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


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YES
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 proven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 15 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) QDPOrderProof [EQUIVALENT, 31 ms]
        (18) QDP
        (19) DependencyGraphProof [EQUIVALENT, 0 ms]
        (20) TRUE


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

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

   rev(nil) -> nil
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)
   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

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:

   rev(nil) -> nil
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)
   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))


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

(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:

   REV(cons(x, l)) -> REV1(x, l)
   REV(cons(x, l)) -> REV2(x, l)
   REV1(x, cons(y, l)) -> REV1(y, l)
   REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
   REV2(x, cons(y, l)) -> REV2(y, l)

The TRS R consists of the following rules:

   rev(nil) -> nil
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)
   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   REV1(x, cons(y, l)) -> REV1(y, l)

The TRS R consists of the following rules:

   rev(nil) -> nil
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)
   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))

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:

   REV1(x, cons(y, l)) -> REV1(y, l)

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

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))

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].

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))


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

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

   REV1(x, cons(y, l)) -> REV1(y, 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:
*REV1(x, cons(y, l)) -> REV1(y, l)
The graph contains the following edges 2 > 1, 2 > 2


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

(13)
YES

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

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

   REV(cons(x, l)) -> REV2(x, l)
   REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
   REV2(x, cons(y, l)) -> REV2(y, l)

The TRS R consists of the following rules:

   rev(nil) -> nil
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)
   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))

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:

   REV(cons(x, l)) -> REV2(x, l)
   REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
   REV2(x, cons(y, l)) -> REV2(y, l)

The TRS R consists of the following rules:

   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)

The set Q consists of the following terms:

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))

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

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


The following pairs can be oriented strictly and are deleted.

   REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
   REV2(x, cons(y, l)) -> REV2(y, l)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( REV_1(x_1) ) = 2x_1
POL( cons_2(x_1, x_2) ) = 2x_2 + 1
POL( rev2_2(x_1, x_2) ) = x_2
POL( nil ) = 0
POL( rev_1(x_1) ) = x_1
POL( rev1_2(x_1, x_2) ) = 0
POL( 0 ) = 2
POL( s_1(x_1) ) = 2x_1 + 2
POL( REV2_2(x_1, x_2) ) = 2x_2 + 2

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

   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))


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

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

   REV(cons(x, l)) -> REV2(x, l)

The TRS R consists of the following rules:

   rev2(x, nil) -> nil
   rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
   rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
   rev1(0, nil) -> 0
   rev1(s(x), nil) -> s(x)
   rev1(x, cons(y, l)) -> rev1(y, l)

The set Q consists of the following terms:

   rev(nil)
   rev(cons(x0, x1))
   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev2(x0, nil)
   rev2(x0, cons(x1, x2))

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

(19) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(20)
TRUE