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


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


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 75 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 29 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 1 ms]
        (21) QDP
        (22) TransformationProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) TransformationProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) TransformationProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) DependencyGraphProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) TransformationProof [EQUIVALENT, 3 ms]
        (35) QDP
        (36) TransformationProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QDPOrderProof [EQUIVALENT, 81 ms]
        (39) QDP
        (40) QDPOrderProof [EQUIVALENT, 83 ms]
        (41) QDP
        (42) SemLabProof [SOUND, 1706 ms]
        (43) QDP
        (44) DependencyGraphProof [EQUIVALENT, 0 ms]
        (45) AND
            (46) QDP
                (47) UsableRulesReductionPairsProof [EQUIVALENT, 5 ms]
                (48) QDP
                (49) MRRProof [EQUIVALENT, 0 ms]
                (50) QDP
                (51) PisEmptyProof [EQUIVALENT, 0 ms]
                (52) YES
            (53) QDP
                (54) MRRProof [EQUIVALENT, 3 ms]
                (55) QDP
                (56) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                (57) QDP
                (58) MRRProof [EQUIVALENT, 0 ms]
                (59) QDP
                (60) DependencyGraphProof [EQUIVALENT, 0 ms]
                (61) QDP
                (62) MRRProof [EQUIVALENT, 0 ms]
                (63) QDP
                (64) DependencyGraphProof [EQUIVALENT, 0 ms]
                (65) TRUE
            (66) QDP
                (67) MRRProof [EQUIVALENT, 0 ms]
                (68) QDP
                (69) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                (70) QDP
                (71) MRRProof [EQUIVALENT, 0 ms]
                (72) QDP
                (73) DependencyGraphProof [EQUIVALENT, 0 ms]
                (74) TRUE


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

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

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   old(free(x)) -> free(old(x))
   new(serve) -> free(serve)
   old(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

Q is empty.

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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(check(x_1)) = x_1
   POL(free(x_1)) = x_1
   POL(new(x_1)) = x_1
   POL(old(x_1)) = 2*x_1
   POL(serve) = 2
   POL(top1(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(top2(x_1, x_2)) = 2*x_1 + 2*x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   old(serve) -> free(serve)




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

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

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   old(free(x)) -> free(old(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

Q is empty.

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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(check(x_1)) = x_1
   POL(free(x_1)) = 2 + x_1
   POL(new(x_1)) = 2 + x_1
   POL(old(x_1)) = 2*x_1
   POL(serve) = 0
   POL(top1(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(top2(x_1, x_2)) = 2*x_1 + 2*x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   old(free(x)) -> free(old(x))




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

Q is empty.

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

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

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

   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP1(free(x), y) -> CHECK(new(x))
   TOP1(free(x), y) -> NEW(x)
   TOP1(free(x), y) -> TOP2(new(x), check(y))
   TOP1(free(x), y) -> CHECK(y)
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP1(free(x), y) -> CHECK(x)
   TOP1(free(x), y) -> NEW(y)
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP1(free(x), y) -> CHECK(new(y))
   TOP2(x, free(y)) -> TOP1(check(new(x)), y)
   TOP2(x, free(y)) -> CHECK(new(x))
   TOP2(x, free(y)) -> NEW(x)
   TOP2(x, free(y)) -> TOP1(new(x), check(y))
   TOP2(x, free(y)) -> CHECK(y)
   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP2(x, free(y)) -> CHECK(x)
   TOP2(x, free(y)) -> NEW(y)
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP2(x, free(y)) -> CHECK(new(y))
   NEW(free(x)) -> NEW(x)
   CHECK(free(x)) -> CHECK(x)
   CHECK(new(x)) -> NEW(check(x))
   CHECK(new(x)) -> CHECK(x)
   CHECK(old(x)) -> CHECK(x)

The TRS R consists of the following rules:

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(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 3 SCCs with 13 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

   NEW(free(x)) -> NEW(x)

The TRS R consists of the following rules:

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(10) 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.
----------------------------------------

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

   NEW(free(x)) -> NEW(x)

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:
*NEW(free(x)) -> NEW(x)
The graph contains the following edges 1 > 1


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

(13)
YES

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

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

   CHECK(new(x)) -> CHECK(x)
   CHECK(free(x)) -> CHECK(x)
   CHECK(old(x)) -> CHECK(x)

The TRS R consists of the following rules:

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(15) 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.
----------------------------------------

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

   CHECK(new(x)) -> CHECK(x)
   CHECK(free(x)) -> CHECK(x)
   CHECK(old(x)) -> CHECK(x)

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

(17) 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:
*CHECK(new(x)) -> CHECK(x)
The graph contains the following edges 1 > 1


*CHECK(free(x)) -> CHECK(x)
The graph contains the following edges 1 > 1


*CHECK(old(x)) -> CHECK(x)
The graph contains the following edges 1 > 1


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

(18)
YES

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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2(x, free(y)) -> TOP1(check(new(x)), y)
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x, free(y)) -> TOP1(new(x), check(y))
   TOP1(free(x), y) -> TOP2(new(x), check(y))
   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(x, check(new(y)))

The TRS R consists of the following rules:

   top1(free(x), y) -> top2(check(new(x)), y)
   top1(free(x), y) -> top2(new(x), check(y))
   top1(free(x), y) -> top2(check(x), new(y))
   top1(free(x), y) -> top2(x, check(new(y)))
   top2(x, free(y)) -> top1(check(new(x)), y)
   top2(x, free(y)) -> top1(new(x), check(y))
   top2(x, free(y)) -> top1(check(x), new(y))
   top2(x, free(y)) -> top1(x, check(new(y)))
   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(20) 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.
----------------------------------------

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

   TOP2(x, free(y)) -> TOP1(check(new(x)), y)
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x, free(y)) -> TOP1(new(x), check(y))
   TOP1(free(x), y) -> TOP2(new(x), check(y))
   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(x, check(new(y)))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(22) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP2(x, free(y)) -> TOP1(check(new(x)), y) at position [0] we obtained the following new rules [LPAR04]:

   (TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1),TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1))
   (TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1),TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1))
   (TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1),TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1))


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

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

   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x, free(y)) -> TOP1(new(x), check(y))
   TOP1(free(x), y) -> TOP2(new(x), check(y))
   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(24) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP2(x, free(y)) -> TOP1(new(x), check(y)) at position [0] we obtained the following new rules [LPAR04]:

   (TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1)),TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1)))
   (TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1)),TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1)))


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

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

   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP1(free(x), y) -> TOP2(new(x), check(y))
   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(26) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP1(free(x), y) -> TOP2(new(x), check(y)) at position [1] we obtained the following new rules [LPAR04]:

   (TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0))),TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0))))
   (TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0))),TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0))))
   (TOP1(free(y0), old(x0)) -> TOP2(new(y0), old(check(x0))),TOP1(free(y0), old(x0)) -> TOP2(new(y0), old(check(x0))))
   (TOP1(free(y0), old(x0)) -> TOP2(new(y0), old(x0)),TOP1(free(y0), old(x0)) -> TOP2(new(y0), old(x0)))


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

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

   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP1(free(y0), old(x0)) -> TOP2(new(y0), old(check(x0)))
   TOP1(free(y0), old(x0)) -> TOP2(new(y0), old(x0))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(28) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   TOP2(x, free(y)) -> TOP1(check(x), new(y))
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(30) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP2(x, free(y)) -> TOP1(check(x), new(y)) at position [0] we obtained the following new rules [LPAR04]:

   (TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1)),TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1)))
   (TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1)),TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1)))
   (TOP2(old(x0), free(y1)) -> TOP1(old(check(x0)), new(y1)),TOP2(old(x0), free(y1)) -> TOP1(old(check(x0)), new(y1)))
   (TOP2(old(x0), free(y1)) -> TOP1(old(x0), new(y1)),TOP2(old(x0), free(y1)) -> TOP1(old(x0), new(y1)))


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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))
   TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1))
   TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1))
   TOP2(old(x0), free(y1)) -> TOP1(old(check(x0)), new(y1))
   TOP2(old(x0), free(y1)) -> TOP1(old(x0), new(y1))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(32) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP1(free(x), y) -> TOP2(check(x), new(y))
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1))
   TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(34) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP1(free(x), y) -> TOP2(check(x), new(y)) at position [1] we obtained the following new rules [LPAR04]:

   (TOP1(free(y0), free(x0)) -> TOP2(check(y0), free(new(x0))),TOP1(free(y0), free(x0)) -> TOP2(check(y0), free(new(x0))))
   (TOP1(free(y0), serve) -> TOP2(check(y0), free(serve)),TOP1(free(y0), serve) -> TOP2(check(y0), free(serve)))


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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP1(free(x), y) -> TOP2(x, check(new(y)))
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1))
   TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))
   TOP1(free(y0), free(x0)) -> TOP2(check(y0), free(new(x0)))
   TOP1(free(y0), serve) -> TOP2(check(y0), free(serve))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(36) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TOP1(free(x), y) -> TOP2(x, check(new(y))) at position [1] we obtained the following new rules [LPAR04]:

   (TOP1(free(y0), x0) -> TOP2(y0, new(check(x0))),TOP1(free(y0), x0) -> TOP2(y0, new(check(x0))))
   (TOP1(free(y0), free(x0)) -> TOP2(y0, check(free(new(x0)))),TOP1(free(y0), free(x0)) -> TOP2(y0, check(free(new(x0)))))
   (TOP1(free(y0), serve) -> TOP2(y0, check(free(serve))),TOP1(free(y0), serve) -> TOP2(y0, check(free(serve))))


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

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1))
   TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))
   TOP1(free(y0), free(x0)) -> TOP2(check(y0), free(new(x0)))
   TOP1(free(y0), serve) -> TOP2(check(y0), free(serve))
   TOP1(free(y0), x0) -> TOP2(y0, new(check(x0)))
   TOP1(free(y0), free(x0)) -> TOP2(y0, check(free(new(x0))))
   TOP1(free(y0), serve) -> TOP2(y0, check(free(serve)))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

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


The following pairs can be oriented strictly and are deleted.

   TOP2(serve, free(y1)) -> TOP1(check(free(serve)), y1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( TOP1_2(x_1, x_2) ) = x_1
POL( TOP2_2(x_1, x_2) ) = x_1
POL( new_1(x_1) ) = x_1
POL( free_1(x_1) ) = x_1
POL( serve ) = 1
POL( check_1(x_1) ) = max{0, -2}
POL( old_1(x_1) ) = max{0, -2}

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

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)


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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1))
   TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))
   TOP1(free(y0), free(x0)) -> TOP2(check(y0), free(new(x0)))
   TOP1(free(y0), serve) -> TOP2(check(y0), free(serve))
   TOP1(free(y0), x0) -> TOP2(y0, new(check(x0)))
   TOP1(free(y0), free(x0)) -> TOP2(y0, check(free(new(x0))))
   TOP1(free(y0), serve) -> TOP2(y0, check(free(serve)))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

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


The following pairs can be oriented strictly and are deleted.

   TOP1(free(y0), serve) -> TOP2(y0, check(free(serve)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( TOP1_2(x_1, x_2) ) = 2x_2
POL( TOP2_2(x_1, x_2) ) = 2x_2
POL( new_1(x_1) ) = x_1
POL( free_1(x_1) ) = x_1
POL( serve ) = 2
POL( check_1(x_1) ) = max{0, -2}
POL( old_1(x_1) ) = max{0, -2}

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

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)


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

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2(x, free(y)) -> TOP1(x, check(new(y)))
   TOP1(free(x), y) -> TOP2(check(new(x)), y)
   TOP2(x0, free(y1)) -> TOP1(new(check(x0)), y1)
   TOP2(free(x0), free(y1)) -> TOP1(check(free(new(x0))), y1)
   TOP1(free(y0), free(x0)) -> TOP2(new(y0), free(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(new(x0)), check(y1))
   TOP1(free(y0), new(x0)) -> TOP2(new(y0), new(check(x0)))
   TOP2(free(x0), free(y1)) -> TOP1(free(check(x0)), new(y1))
   TOP2(new(x0), free(y1)) -> TOP1(new(check(x0)), new(y1))
   TOP2(serve, free(y1)) -> TOP1(free(serve), check(y1))
   TOP1(free(y0), free(x0)) -> TOP2(check(y0), free(new(x0)))
   TOP1(free(y0), serve) -> TOP2(check(y0), free(serve))
   TOP1(free(y0), x0) -> TOP2(y0, new(check(x0)))
   TOP1(free(y0), free(x0)) -> TOP2(y0, check(free(new(x0))))

The TRS R consists of the following rules:

   new(free(x)) -> free(new(x))
   new(serve) -> free(serve)
   check(free(x)) -> free(check(x))
   check(new(x)) -> new(check(x))
   check(old(x)) -> old(check(x))
   check(old(x)) -> old(x)

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

(42) SemLabProof (SOUND)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
old: 1
new: x<sub>0</sub>
TOP2: 0
serve: 0
check: 1
free: x<sub>0</sub>
TOP1: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.0-0(x, free.0(y)) -> TOP1.0-1(x, check.0(new.0(y)))
   TOP1.0-0(free.0(x), y) -> TOP2.1-0(check.0(new.0(x)), y)
   TOP1.0-1(free.0(x), y) -> TOP2.1-1(check.0(new.0(x)), y)
   TOP1.1-0(free.1(x), y) -> TOP2.1-0(check.1(new.1(x)), y)
   TOP1.1-1(free.1(x), y) -> TOP2.1-1(check.1(new.1(x)), y)
   TOP2.0-1(x, free.1(y)) -> TOP1.0-1(x, check.1(new.1(y)))
   TOP2.1-0(x, free.0(y)) -> TOP1.1-1(x, check.0(new.0(y)))
   TOP2.1-1(x, free.1(y)) -> TOP1.1-1(x, check.1(new.1(y)))
   TOP1.0-0(free.0(y0), free.0(x0)) -> TOP2.0-1(new.0(y0), free.1(check.0(x0)))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(new.0(y0), free.1(check.1(x0)))
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-1(new.1(y0), free.1(check.0(x0)))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(new.1(y0), free.1(check.1(x0)))
   TOP1.0-0(free.0(y0), new.0(x0)) -> TOP2.0-1(new.0(y0), new.1(check.0(x0)))
   TOP1.0-1(free.0(y0), new.1(x0)) -> TOP2.0-1(new.0(y0), new.1(check.1(x0)))
   TOP1.1-0(free.1(y0), new.0(x0)) -> TOP2.1-1(new.1(y0), new.1(check.0(x0)))
   TOP1.1-1(free.1(y0), new.1(x0)) -> TOP2.1-1(new.1(y0), new.1(check.1(x0)))
   TOP1.0-0(free.0(y0), free.0(x0)) -> TOP2.1-0(check.0(y0), free.0(new.0(x0)))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.1-1(check.0(y0), free.1(new.1(x0)))
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(check.1(y0), free.1(new.1(x0)))
   TOP1.0-0(free.0(y0), x0) -> TOP2.0-1(y0, new.1(check.0(x0)))
   TOP1.0-1(free.0(y0), x0) -> TOP2.0-1(y0, new.1(check.1(x0)))
   TOP1.1-0(free.1(y0), x0) -> TOP2.1-1(y0, new.1(check.0(x0)))
   TOP1.1-1(free.1(y0), x0) -> TOP2.1-1(y0, new.1(check.1(x0)))
   TOP1.0-0(free.0(y0), free.0(x0)) -> TOP2.0-1(y0, check.0(free.0(new.0(x0))))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(y0, check.1(free.1(new.1(x0))))
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-1(y0, check.0(free.0(new.0(x0))))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(y0, check.1(free.1(new.1(x0))))
   TOP2.0-0(x0, free.0(y1)) -> TOP1.1-0(new.1(check.0(x0)), y1)
   TOP2.0-1(x0, free.1(y1)) -> TOP1.1-1(new.1(check.0(x0)), y1)
   TOP2.1-0(x0, free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), y1)
   TOP2.1-1(x0, free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), y1)
   TOP2.0-0(free.0(x0), free.0(y1)) -> TOP1.1-0(check.0(free.0(new.0(x0))), y1)
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.1-1(check.0(free.0(new.0(x0))), y1)
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(check.1(free.1(new.1(x0))), y1)
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(check.1(free.1(new.1(x0))), y1)
   TOP2.0-0(free.0(x0), free.0(y1)) -> TOP1.0-1(free.0(new.0(x0)), check.0(y1))
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.0-1(free.0(new.0(x0)), check.1(y1))
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-1(free.1(new.1(x0)), check.0(y1))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(new.1(x0)), check.1(y1))
   TOP2.0-0(free.0(x0), free.0(y1)) -> TOP1.1-0(free.1(check.0(x0)), new.0(y1))
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.1-1(free.1(check.0(x0)), new.1(y1))
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(check.1(x0)), new.1(y1))
   TOP2.0-0(new.0(x0), free.0(y1)) -> TOP1.1-0(new.1(check.0(x0)), new.0(y1))
   TOP2.0-1(new.0(x0), free.1(y1)) -> TOP1.1-1(new.1(check.0(x0)), new.1(y1))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))
   TOP2.1-1(new.1(x0), free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), new.1(y1))
   TOP1.0-0(free.0(y0), serve.) -> TOP2.1-0(check.0(y0), free.0(serve.))
   TOP1.1-0(free.1(y0), serve.) -> TOP2.1-0(check.1(y0), free.0(serve.))
   TOP2.0-0(serve., free.0(y1)) -> TOP1.0-1(free.0(serve.), check.0(y1))
   TOP2.0-1(serve., free.1(y1)) -> TOP1.0-1(free.0(serve.), check.1(y1))

The TRS R consists of the following rules:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.0(x)) -> old.1(check.0(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(44) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 26 less nodes.
----------------------------------------

(45)
Complex Obligation (AND)

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

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.1-1(x, free.1(y)) -> TOP1.1-1(x, check.1(new.1(y)))
   TOP1.1-1(free.1(x), y) -> TOP2.1-1(check.1(new.1(x)), y)
   TOP2.1-1(x0, free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), y1)
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(new.1(y0), free.1(check.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(check.1(free.1(new.1(x0))), y1)
   TOP1.1-1(free.1(y0), new.1(x0)) -> TOP2.1-1(new.1(y0), new.1(check.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(new.1(x0)), check.1(y1))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(check.1(y0), free.1(new.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(check.1(x0)), new.1(y1))
   TOP1.1-1(free.1(y0), x0) -> TOP2.1-1(y0, new.1(check.1(x0)))
   TOP2.1-1(new.1(x0), free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), new.1(y1))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(y0, check.1(free.1(new.1(x0))))

The TRS R consists of the following rules:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.0(x)) -> old.1(check.0(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(47) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(old.0(x)) -> old.1(check.0(x))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(TOP1.1-1(x_1, x_2)) = x_1 + x_2
   POL(TOP2.1-1(x_1, x_2)) = x_1 + x_2
   POL(check.0(x_1)) = x_1
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = 1 + x_1
   POL(free.1(x_1)) = 1 + x_1
   POL(new.0(x_1)) = 1 + x_1
   POL(new.1(x_1)) = 1 + x_1
   POL(old.0(x_1)) = 1 + x_1
   POL(old.1(x_1)) = x_1


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

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.1-1(x, free.1(y)) -> TOP1.1-1(x, check.1(new.1(y)))
   TOP1.1-1(free.1(x), y) -> TOP2.1-1(check.1(new.1(x)), y)
   TOP2.1-1(x0, free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), y1)
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(new.1(y0), free.1(check.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(check.1(free.1(new.1(x0))), y1)
   TOP1.1-1(free.1(y0), new.1(x0)) -> TOP2.1-1(new.1(y0), new.1(check.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(new.1(x0)), check.1(y1))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(check.1(y0), free.1(new.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(check.1(x0)), new.1(y1))
   TOP1.1-1(free.1(y0), x0) -> TOP2.1-1(y0, new.1(check.1(x0)))
   TOP2.1-1(new.1(x0), free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), new.1(y1))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(y0, check.1(free.1(new.1(x0))))

The TRS R consists of the following rules:

   new.1(free.1(x)) -> free.1(new.1(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(49) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   TOP2.1-1(x, free.1(y)) -> TOP1.1-1(x, check.1(new.1(y)))
   TOP1.1-1(free.1(x), y) -> TOP2.1-1(check.1(new.1(x)), y)
   TOP2.1-1(x0, free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), y1)
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(new.1(y0), free.1(check.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(check.1(free.1(new.1(x0))), y1)
   TOP1.1-1(free.1(y0), new.1(x0)) -> TOP2.1-1(new.1(y0), new.1(check.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(new.1(x0)), check.1(y1))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(check.1(y0), free.1(new.1(x0)))
   TOP2.1-1(free.1(x0), free.1(y1)) -> TOP1.1-1(free.1(check.1(x0)), new.1(y1))
   TOP1.1-1(free.1(y0), x0) -> TOP2.1-1(y0, new.1(check.1(x0)))
   TOP2.1-1(new.1(x0), free.1(y1)) -> TOP1.1-1(new.1(check.1(x0)), new.1(y1))
   TOP1.1-1(free.1(y0), free.1(x0)) -> TOP2.1-1(y0, check.1(free.1(new.1(x0))))


Used ordering: Polynomial interpretation [POLO]:

   POL(TOP1.1-1(x_1, x_2)) = x_1 + x_2
   POL(TOP2.1-1(x_1, x_2)) = x_1 + x_2
   POL(check.1(x_1)) = x_1
   POL(free.1(x_1)) = 1 + x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = x_1
   POL(old.1(x_1)) = x_1


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

(50)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   new.1(free.1(x)) -> free.1(new.1(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(51) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(52)
YES

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

(53)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.1-0(x0, free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), y1)
   TOP1.1-0(free.1(x), y) -> TOP2.1-0(check.1(new.1(x)), y)
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(check.1(free.1(new.1(x0))), y1)
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))
   TOP1.1-0(free.1(y0), serve.) -> TOP2.1-0(check.1(y0), free.0(serve.))

The TRS R consists of the following rules:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.0(x)) -> old.1(check.0(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(54) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   check.1(old.0(x)) -> old.1(check.0(x))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP1.1-0(x_1, x_2)) = x_1 + x_2
   POL(TOP2.1-0(x_1, x_2)) = x_1 + x_2
   POL(check.0(x_1)) = x_1
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = x_1
   POL(free.1(x_1)) = x_1
   POL(new.0(x_1)) = x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = 1 + x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(55)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.1-0(x0, free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), y1)
   TOP1.1-0(free.1(x), y) -> TOP2.1-0(check.1(new.1(x)), y)
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(check.1(free.1(new.1(x0))), y1)
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))
   TOP1.1-0(free.1(y0), serve.) -> TOP2.1-0(check.1(y0), free.0(serve.))

The TRS R consists of the following rules:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(56) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   check.0(free.0(x)) -> free.1(check.0(x))
   check.0(new.0(x)) -> new.1(check.0(x))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(TOP1.1-0(x_1, x_2)) = x_1 + x_2
   POL(TOP2.1-0(x_1, x_2)) = x_1 + x_2
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = x_1
   POL(free.1(x_1)) = x_1
   POL(new.0(x_1)) = x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(57)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.1-0(x0, free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), y1)
   TOP1.1-0(free.1(x), y) -> TOP2.1-0(check.1(new.1(x)), y)
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(check.1(free.1(new.1(x0))), y1)
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))
   TOP1.1-0(free.1(y0), serve.) -> TOP2.1-0(check.1(y0), free.0(serve.))

The TRS R consists of the following rules:

   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)

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

(58) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   TOP2.1-0(x0, free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), y1)
   TOP1.1-0(free.1(x), y) -> TOP2.1-0(check.1(new.1(x)), y)
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(check.1(free.1(new.1(x0))), y1)


Used ordering: Polynomial interpretation [POLO]:

   POL(TOP1.1-0(x_1, x_2)) = x_1 + x_2
   POL(TOP2.1-0(x_1, x_2)) = x_1 + x_2
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = 1 + x_1
   POL(free.1(x_1)) = 1 + x_1
   POL(new.0(x_1)) = 1 + x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(59)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))
   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))
   TOP1.1-0(free.1(y0), serve.) -> TOP2.1-0(check.1(y0), free.0(serve.))

The TRS R consists of the following rules:

   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)

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

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

(61)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))

The TRS R consists of the following rules:

   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)

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

(62) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   TOP2.1-0(free.1(x0), free.0(y1)) -> TOP1.1-0(free.1(check.1(x0)), new.0(y1))
   TOP2.1-0(new.1(x0), free.0(y1)) -> TOP1.1-0(new.1(check.1(x0)), new.0(y1))


Used ordering: Polynomial interpretation [POLO]:

   POL(TOP1.1-0(x_1, x_2)) = x_1 + x_2
   POL(TOP2.1-0(x_1, x_2)) = 1 + x_1 + x_2
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = x_1
   POL(free.1(x_1)) = 1 + x_1
   POL(new.0(x_1)) = x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(63)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP1.1-0(free.1(y0), free.0(x0)) -> TOP2.1-0(check.1(y0), free.0(new.0(x0)))

The TRS R consists of the following rules:

   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)

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

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

(65)
TRUE

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

(66)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(new.0(y0), free.1(check.1(x0)))
   TOP2.0-1(x, free.1(y)) -> TOP1.0-1(x, check.1(new.1(y)))
   TOP1.0-1(free.0(y0), new.1(x0)) -> TOP2.0-1(new.0(y0), new.1(check.1(x0)))
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.0-1(free.0(new.0(x0)), check.1(y1))
   TOP1.0-1(free.0(y0), x0) -> TOP2.0-1(y0, new.1(check.1(x0)))
   TOP2.0-1(serve., free.1(y1)) -> TOP1.0-1(free.0(serve.), check.1(y1))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(y0, check.1(free.1(new.1(x0))))

The TRS R consists of the following rules:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.0(x)) -> old.1(check.0(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(67) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   check.1(old.0(x)) -> old.1(check.0(x))

Used ordering: Polynomial interpretation [POLO]:

   POL(TOP1.0-1(x_1, x_2)) = x_1 + x_2
   POL(TOP2.0-1(x_1, x_2)) = x_1 + x_2
   POL(check.0(x_1)) = x_1
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = x_1
   POL(free.1(x_1)) = x_1
   POL(new.0(x_1)) = x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = 1 + x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(68)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(new.0(y0), free.1(check.1(x0)))
   TOP2.0-1(x, free.1(y)) -> TOP1.0-1(x, check.1(new.1(y)))
   TOP1.0-1(free.0(y0), new.1(x0)) -> TOP2.0-1(new.0(y0), new.1(check.1(x0)))
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.0-1(free.0(new.0(x0)), check.1(y1))
   TOP1.0-1(free.0(y0), x0) -> TOP2.0-1(y0, new.1(check.1(x0)))
   TOP2.0-1(serve., free.1(y1)) -> TOP1.0-1(free.0(serve.), check.1(y1))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(y0, check.1(free.1(new.1(x0))))

The TRS R consists of the following rules:

   new.0(free.0(x)) -> free.0(new.0(x))
   new.1(free.1(x)) -> free.1(new.1(x))
   new.0(serve.) -> free.0(serve.)
   check.0(free.0(x)) -> free.1(check.0(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.0(new.0(x)) -> new.1(check.0(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)

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

(69) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   check.0(free.0(x)) -> free.1(check.0(x))
   check.0(new.0(x)) -> new.1(check.0(x))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(TOP1.0-1(x_1, x_2)) = x_1 + x_2
   POL(TOP2.0-1(x_1, x_2)) = x_1 + x_2
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = x_1
   POL(free.1(x_1)) = x_1
   POL(new.0(x_1)) = x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(new.0(y0), free.1(check.1(x0)))
   TOP2.0-1(x, free.1(y)) -> TOP1.0-1(x, check.1(new.1(y)))
   TOP1.0-1(free.0(y0), new.1(x0)) -> TOP2.0-1(new.0(y0), new.1(check.1(x0)))
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.0-1(free.0(new.0(x0)), check.1(y1))
   TOP1.0-1(free.0(y0), x0) -> TOP2.0-1(y0, new.1(check.1(x0)))
   TOP2.0-1(serve., free.1(y1)) -> TOP1.0-1(free.0(serve.), check.1(y1))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(y0, check.1(free.1(new.1(x0))))

The TRS R consists of the following rules:

   new.1(free.1(x)) -> free.1(new.1(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)
   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)

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

(71) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(new.0(y0), free.1(check.1(x0)))
   TOP1.0-1(free.0(y0), new.1(x0)) -> TOP2.0-1(new.0(y0), new.1(check.1(x0)))
   TOP1.0-1(free.0(y0), x0) -> TOP2.0-1(y0, new.1(check.1(x0)))
   TOP1.0-1(free.0(y0), free.1(x0)) -> TOP2.0-1(y0, check.1(free.1(new.1(x0))))


Used ordering: Polynomial interpretation [POLO]:

   POL(TOP1.0-1(x_1, x_2)) = 1 + x_1 + x_2
   POL(TOP2.0-1(x_1, x_2)) = x_1 + x_2
   POL(check.1(x_1)) = x_1
   POL(free.0(x_1)) = x_1
   POL(free.1(x_1)) = 1 + x_1
   POL(new.0(x_1)) = x_1
   POL(new.1(x_1)) = x_1
   POL(old.0(x_1)) = x_1
   POL(old.1(x_1)) = x_1
   POL(serve.) = 0


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

(72)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP2.0-1(x, free.1(y)) -> TOP1.0-1(x, check.1(new.1(y)))
   TOP2.0-1(free.0(x0), free.1(y1)) -> TOP1.0-1(free.0(new.0(x0)), check.1(y1))
   TOP2.0-1(serve., free.1(y1)) -> TOP1.0-1(free.0(serve.), check.1(y1))

The TRS R consists of the following rules:

   new.1(free.1(x)) -> free.1(new.1(x))
   check.1(free.1(x)) -> free.1(check.1(x))
   check.1(new.1(x)) -> new.1(check.1(x))
   check.1(old.1(x)) -> old.1(check.1(x))
   check.1(old.0(x)) -> old.0(x)
   check.1(old.1(x)) -> old.1(x)
   new.0(free.0(x)) -> free.0(new.0(x))
   new.0(serve.) -> free.0(serve.)

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

(73) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
----------------------------------------

(74)
TRUE