/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 13 ms]
(4) QDP
(5) SemLabProof [SOUND, 212 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) PisEmptyProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) PisEmptyProof [EQUIVALENT, 0 ms]
        (28) YES
    (29) QDP
        (30) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) PisEmptyProof [EQUIVALENT, 0 ms]
        (33) YES
    (34) QDP
        (35) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) PisEmptyProof [EQUIVALENT, 0 ms]
        (38) YES
    (39) QDP
        (40) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) PisEmptyProof [EQUIVALENT, 0 ms]
        (43) YES
    (44) QDP
        (45) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) PisEmptyProof [EQUIVALENT, 0 ms]
        (48) YES


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

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

   f(0, 1, x) -> f(g(x), g(x), x)
   f(g(x), y, z) -> g(f(x, y, z))
   f(x, g(y), z) -> g(f(x, y, z))
   f(x, y, g(z)) -> g(f(x, y, z))

Q is empty.

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

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

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

   F(0, 1, x) -> F(g(x), g(x), x)
   F(g(x), y, z) -> F(x, y, z)
   F(x, g(y), z) -> F(x, y, z)
   F(x, y, g(z)) -> F(x, y, z)

The TRS R consists of the following rules:

   f(0, 1, x) -> f(g(x), g(x), x)
   f(g(x), y, z) -> g(f(x, y, z))
   f(x, g(y), z) -> g(f(x, y, z))
   f(x, y, g(z)) -> g(f(x, y, z))

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

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


The following pairs can be oriented strictly and are deleted.

   F(x, y, g(z)) -> F(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1, x2, x3)  =  x3

g(x1)  =  g(x1)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   g_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

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

   F(0, 1, x) -> F(g(x), g(x), x)
   F(g(x), y, z) -> F(x, y, z)
   F(x, g(y), z) -> F(x, y, z)

The TRS R consists of the following rules:

   f(0, 1, x) -> f(g(x), g(x), x)
   f(g(x), y, z) -> g(f(x, y, z))
   f(x, g(y), z) -> g(f(x, y, z))
   f(x, y, g(z)) -> g(f(x, y, z))

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

(5) 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.
f: 0
0: 0
1: 1
F: 0
g: x<sub>0</sub>
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

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

   F.0-1-0(0., 1., x) -> F.0-0-0(g.0(x), g.0(x), x)
   F.0-0-0(g.0(x), y, z) -> F.0-0-0(x, y, z)
   F.0-0-1(g.0(x), y, z) -> F.0-0-1(x, y, z)
   F.0-1-0(g.0(x), y, z) -> F.0-1-0(x, y, z)
   F.0-1-1(g.0(x), y, z) -> F.0-1-1(x, y, z)
   F.1-0-0(g.1(x), y, z) -> F.1-0-0(x, y, z)
   F.1-0-1(g.1(x), y, z) -> F.1-0-1(x, y, z)
   F.1-1-0(g.1(x), y, z) -> F.1-1-0(x, y, z)
   F.1-1-1(g.1(x), y, z) -> F.1-1-1(x, y, z)
   F.0-1-1(0., 1., x) -> F.1-1-1(g.1(x), g.1(x), x)
   F.0-0-0(x, g.0(y), z) -> F.0-0-0(x, y, z)
   F.0-0-1(x, g.0(y), z) -> F.0-0-1(x, y, z)
   F.0-1-0(x, g.1(y), z) -> F.0-1-0(x, y, z)
   F.0-1-1(x, g.1(y), z) -> F.0-1-1(x, y, z)
   F.1-0-0(x, g.0(y), z) -> F.1-0-0(x, y, z)
   F.1-0-1(x, g.0(y), z) -> F.1-0-1(x, y, z)
   F.1-1-0(x, g.1(y), z) -> F.1-1-0(x, y, z)
   F.1-1-1(x, g.1(y), z) -> F.1-1-1(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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 8 SCCs with 2 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

   F.1-1-1(x, g.1(y), z) -> F.1-1-1(x, y, z)
   F.1-1-1(g.1(x), y, z) -> F.1-1-1(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(10) 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.

The following dependency pairs can be deleted:

   F.1-1-1(x, g.1(y), z) -> F.1-1-1(x, y, z)
   F.1-1-1(g.1(x), y, z) -> F.1-1-1(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.1-1-1(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.1(x_1)) = x_1


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

(11)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(13)
YES

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

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

   F.1-1-0(x, g.1(y), z) -> F.1-1-0(x, y, z)
   F.1-1-0(g.1(x), y, z) -> F.1-1-0(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(15) 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.

The following dependency pairs can be deleted:

   F.1-1-0(x, g.1(y), z) -> F.1-1-0(x, y, z)
   F.1-1-0(g.1(x), y, z) -> F.1-1-0(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.1-1-0(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.1(x_1)) = x_1


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

(16)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(18)
YES

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

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

   F.1-0-1(x, g.0(y), z) -> F.1-0-1(x, y, z)
   F.1-0-1(g.1(x), y, z) -> F.1-0-1(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(20) 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.

The following dependency pairs can be deleted:

   F.1-0-1(x, g.0(y), z) -> F.1-0-1(x, y, z)
   F.1-0-1(g.1(x), y, z) -> F.1-0-1(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.1-0-1(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1


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

(21)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(23)
YES

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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F.1-0-0(x, g.0(y), z) -> F.1-0-0(x, y, z)
   F.1-0-0(g.1(x), y, z) -> F.1-0-0(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(25) 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.

The following dependency pairs can be deleted:

   F.1-0-0(x, g.0(y), z) -> F.1-0-0(x, y, z)
   F.1-0-0(g.1(x), y, z) -> F.1-0-0(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.1-0-0(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1


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

(26)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(28)
YES

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

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

   F.0-1-1(x, g.1(y), z) -> F.0-1-1(x, y, z)
   F.0-1-1(g.0(x), y, z) -> F.0-1-1(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(30) 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.

The following dependency pairs can be deleted:

   F.0-1-1(x, g.1(y), z) -> F.0-1-1(x, y, z)
   F.0-1-1(g.0(x), y, z) -> F.0-1-1(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.0-1-1(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1


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

(31)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(33)
YES

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

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

   F.0-0-1(x, g.0(y), z) -> F.0-0-1(x, y, z)
   F.0-0-1(g.0(x), y, z) -> F.0-0-1(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(35) 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.

The following dependency pairs can be deleted:

   F.0-0-1(x, g.0(y), z) -> F.0-0-1(x, y, z)
   F.0-0-1(g.0(x), y, z) -> F.0-0-1(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.0-0-1(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.0(x_1)) = x_1


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

(36)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(38)
YES

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

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

   F.0-0-0(x, g.0(y), z) -> F.0-0-0(x, y, z)
   F.0-0-0(g.0(x), y, z) -> F.0-0-0(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(40) 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.

The following dependency pairs can be deleted:

   F.0-0-0(x, g.0(y), z) -> F.0-0-0(x, y, z)
   F.0-0-0(g.0(x), y, z) -> F.0-0-0(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.0-0-0(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.0(x_1)) = x_1


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

(41)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(43)
YES

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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F.0-1-0(x, g.1(y), z) -> F.0-1-0(x, y, z)
   F.0-1-0(g.0(x), y, z) -> F.0-1-0(x, y, z)

The TRS R consists of the following rules:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))

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

(45) 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.

The following dependency pairs can be deleted:

   F.0-1-0(x, g.1(y), z) -> F.0-1-0(x, y, z)
   F.0-1-0(g.0(x), y, z) -> F.0-1-0(x, y, z)
The following rules are removed from R:

   f.0-1-0(0., 1., x) -> f.0-0-0(g.0(x), g.0(x), x)
   f.0-1-1(0., 1., x) -> f.1-1-1(g.1(x), g.1(x), x)
   f.0-0-0(g.0(x), y, z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(g.0(x), y, z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(g.0(x), y, z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(g.0(x), y, z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(g.1(x), y, z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(g.1(x), y, z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(g.1(x), y, z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(g.1(x), y, z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, g.0(y), z) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, g.0(y), z) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, g.1(y), z) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, g.1(y), z) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, g.0(y), z) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, g.0(y), z) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, g.1(y), z) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, g.1(y), z) -> g.0(f.1-1-1(x, y, z))
   f.0-0-0(x, y, g.0(z)) -> g.0(f.0-0-0(x, y, z))
   f.0-0-1(x, y, g.1(z)) -> g.0(f.0-0-1(x, y, z))
   f.0-1-0(x, y, g.0(z)) -> g.0(f.0-1-0(x, y, z))
   f.0-1-1(x, y, g.1(z)) -> g.0(f.0-1-1(x, y, z))
   f.1-0-0(x, y, g.0(z)) -> g.0(f.1-0-0(x, y, z))
   f.1-0-1(x, y, g.1(z)) -> g.0(f.1-0-1(x, y, z))
   f.1-1-0(x, y, g.0(z)) -> g.0(f.1-1-0(x, y, z))
   f.1-1-1(x, y, g.1(z)) -> g.0(f.1-1-1(x, y, z))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(F.0-1-0(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(g.0(x_1)) = x_1
   POL(g.1(x_1)) = x_1


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

(46)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(47) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(48)
YES