/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 27 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) TransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) TransformationProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) DependencyGraphProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) TransformationProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) DependencyGraphProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) SplitQDPProof [EQUIVALENT, 0 ms]
        (23) AND
            (24) QDP
                (25) SemLabProof [SOUND, 0 ms]
                (26) QDP
                (27) DependencyGraphProof [EQUIVALENT, 0 ms]
                (28) QDP
                (29) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                (30) QDP
                (31) MRRProof [EQUIVALENT, 0 ms]
                (32) QDP
                (33) PisEmptyProof [SOUND, 0 ms]
                (34) TRUE
            (35) QDP
                (36) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
                (37) QDP
                (38) RFCMatchBoundsDPProof [EQUIVALENT, 0 ms]
                (39) YES
    (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) QDPOrderProof [EQUIVALENT, 232 ms]
        (44) QDP
        (45) PisEmptyProof [EQUIVALENT, 0 ms]
        (46) YES


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

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

   g(x, x) -> g(a, b)
   g(c, g(c, x)) -> g(e, g(d, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(e, g(e, x)) -> g(d, g(c, x))
   f(g(x, y)) -> g(y, g(f(f(x)), a))

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:

   G(x, x) -> G(a, b)
   G(c, g(c, x)) -> G(e, g(d, x))
   G(c, g(c, x)) -> G(d, x)
   G(d, g(d, x)) -> G(c, g(e, x))
   G(d, g(d, x)) -> G(e, x)
   G(e, g(e, x)) -> G(d, g(c, x))
   G(e, g(e, x)) -> G(c, x)
   F(g(x, y)) -> G(y, g(f(f(x)), a))
   F(g(x, y)) -> G(f(f(x)), a)
   F(g(x, y)) -> F(f(x))
   F(g(x, y)) -> F(x)

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(c, g(c, x)) -> g(e, g(d, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(e, g(e, x)) -> g(d, g(c, x))
   f(g(x, y)) -> g(y, g(f(f(x)), a))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(e, g(e, x)) -> G(d, g(c, x))
   G(d, g(d, x)) -> G(c, g(e, x))
   G(c, g(c, x)) -> G(e, g(d, x))
   G(e, g(e, x)) -> G(c, x)
   G(c, g(c, x)) -> G(d, x)
   G(d, g(d, x)) -> G(e, x)

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(c, g(c, x)) -> g(e, g(d, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(e, g(e, x)) -> g(d, g(c, x))
   f(g(x, y)) -> g(y, g(f(f(x)), a))

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

(6) 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.
----------------------------------------

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

   G(e, g(e, x)) -> G(d, g(c, x))
   G(d, g(d, x)) -> G(c, g(e, x))
   G(c, g(c, x)) -> G(e, g(d, x))
   G(e, g(e, x)) -> G(c, x)
   G(c, g(c, x)) -> G(d, x)
   G(d, g(d, x)) -> G(e, x)

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

(8) 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:

   G(e, g(e, x)) -> G(c, x)
   G(c, g(c, x)) -> G(d, x)
   G(d, g(d, x)) -> G(e, x)


Used ordering: Polynomial interpretation [POLO]:

   POL(G(x_1, x_2)) = x_1 + 2*x_2
   POL(a) = 0
   POL(b) = 0
   POL(c) = 0
   POL(d) = 0
   POL(e) = 0
   POL(g(x_1, x_2)) = 1 + 2*x_1 + 2*x_2


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

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

   G(e, g(e, x)) -> G(d, g(c, x))
   G(d, g(d, x)) -> G(c, g(e, x))
   G(c, g(c, x)) -> G(e, g(d, x))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

(10) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule G(e, g(e, x)) -> G(d, g(c, x)) at position [1] we obtained the following new rules [LPAR04]:

   (G(e, g(e, c)) -> G(d, g(a, b)),G(e, g(e, c)) -> G(d, g(a, b)))
   (G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0))),G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0))))


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

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

   G(d, g(d, x)) -> G(c, g(e, x))
   G(c, g(c, x)) -> G(e, g(d, x))
   G(e, g(e, c)) -> G(d, g(a, b))
   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(c, g(c, x)) -> G(e, g(d, x))
   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))
   G(d, g(d, x)) -> G(c, g(e, x))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

(14) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule G(c, g(c, x)) -> G(e, g(d, x)) at position [1] we obtained the following new rules [LPAR04]:

   (G(c, g(c, d)) -> G(e, g(a, b)),G(c, g(c, d)) -> G(e, g(a, b)))
   (G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0))),G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0))))


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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))
   G(d, g(d, x)) -> G(c, g(e, x))
   G(c, g(c, d)) -> G(e, g(a, b))
   G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

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

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G(d, g(d, x)) -> G(c, g(e, x))
   G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0)))
   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

(18) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule G(d, g(d, x)) -> G(c, g(e, x)) at position [1] we obtained the following new rules [LPAR04]:

   (G(d, g(d, e)) -> G(c, g(a, b)),G(d, g(d, e)) -> G(c, g(a, b)))
   (G(d, g(d, g(e, x0))) -> G(c, g(d, g(c, x0))),G(d, g(d, g(e, x0))) -> G(c, g(d, g(c, x0))))


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

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

   G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0)))
   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))
   G(d, g(d, e)) -> G(c, g(a, b))
   G(d, g(d, g(e, x0))) -> G(c, g(d, g(c, x0)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

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

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

   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))
   G(d, g(d, g(e, x0))) -> G(c, g(d, g(c, x0)))
   G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

(22) SplitQDPProof (EQUIVALENT)
We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem

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

(23)
Complex Obligation (AND)

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

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

   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))
   G(d, g(d, g(e, x0))) -> G(c, g(d, g(c, x0)))
   G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))

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

(25) 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.
a: 0
b: 0
c: 1
G: 0
d: 1
e: 1
g: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   G.1-0(e., g.1-0(e., g.1-0(c., x0))) -> G.1-0(d., g.1-0(e., g.1-0(d., x0)))
   G.1-0(d., g.1-0(d., g.1-0(e., x0))) -> G.1-0(c., g.1-0(d., g.1-0(c., x0)))
   G.1-0(d., g.1-0(d., g.1-1(e., x0))) -> G.1-0(c., g.1-0(d., g.1-1(c., x0)))
   G.1-0(e., g.1-0(e., g.1-1(c., x0))) -> G.1-0(d., g.1-0(e., g.1-1(d., x0)))
   G.1-0(c., g.1-0(c., g.1-0(d., x0))) -> G.1-0(e., g.1-0(c., g.1-0(e., x0)))
   G.1-0(c., g.1-0(c., g.1-1(d., x0))) -> G.1-0(e., g.1-0(c., g.1-1(e., x0)))

The TRS R consists of the following rules:

   g.0-0(x, x) -> g.0-0(a., b.)
   g.1-1(x, x) -> g.0-0(a., b.)
   g.1-0(e., g.1-0(e., x)) -> g.1-0(d., g.1-0(c., x))
   g.1-0(e., g.1-1(e., x)) -> g.1-0(d., g.1-1(c., x))
   g.1-0(d., g.1-0(d., x)) -> g.1-0(c., g.1-0(e., x))
   g.1-0(d., g.1-1(d., x)) -> g.1-0(c., g.1-1(e., x))
   g.1-0(c., g.1-0(c., x)) -> g.1-0(e., g.1-0(d., x))
   g.1-0(c., g.1-1(c., x)) -> g.1-0(e., g.1-1(d., x))

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

(27) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

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

   G.1-0(d., g.1-0(d., g.1-0(e., x0))) -> G.1-0(c., g.1-0(d., g.1-0(c., x0)))
   G.1-0(c., g.1-0(c., g.1-0(d., x0))) -> G.1-0(e., g.1-0(c., g.1-0(e., x0)))
   G.1-0(e., g.1-0(e., g.1-0(c., x0))) -> G.1-0(d., g.1-0(e., g.1-0(d., x0)))

The TRS R consists of the following rules:

   g.0-0(x, x) -> g.0-0(a., b.)
   g.1-1(x, x) -> g.0-0(a., b.)
   g.1-0(e., g.1-0(e., x)) -> g.1-0(d., g.1-0(c., x))
   g.1-0(e., g.1-1(e., x)) -> g.1-0(d., g.1-1(c., x))
   g.1-0(d., g.1-0(d., x)) -> g.1-0(c., g.1-0(e., x))
   g.1-0(d., g.1-1(d., x)) -> g.1-0(c., g.1-1(e., x))
   g.1-0(c., g.1-0(c., x)) -> g.1-0(e., g.1-0(d., x))
   g.1-0(c., g.1-1(c., x)) -> g.1-0(e., g.1-1(d., x))

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

(29) 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:

   g.0-0(x, x) -> g.0-0(a., b.)
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(G.1-0(x_1, x_2)) = x_1 + x_2
   POL(a.) = 1
   POL(b.) = 0
   POL(c.) = 1
   POL(d.) = 1
   POL(e.) = 1
   POL(g.0-0(x_1, x_2)) = x_1 + x_2
   POL(g.1-0(x_1, x_2)) = x_1 + x_2
   POL(g.1-1(x_1, x_2)) = 1 + x_1 + x_2


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

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

   G.1-0(d., g.1-0(d., g.1-0(e., x0))) -> G.1-0(c., g.1-0(d., g.1-0(c., x0)))
   G.1-0(c., g.1-0(c., g.1-0(d., x0))) -> G.1-0(e., g.1-0(c., g.1-0(e., x0)))
   G.1-0(e., g.1-0(e., g.1-0(c., x0))) -> G.1-0(d., g.1-0(e., g.1-0(d., x0)))

The TRS R consists of the following rules:

   g.1-0(d., g.1-0(d., x)) -> g.1-0(c., g.1-0(e., x))
   g.1-0(c., g.1-0(c., x)) -> g.1-0(e., g.1-0(d., x))
   g.1-0(e., g.1-0(e., x)) -> g.1-0(d., g.1-0(c., x))
   g.1-0(d., g.1-1(d., x)) -> g.1-0(c., g.1-1(e., x))
   g.1-0(c., g.1-1(c., x)) -> g.1-0(e., g.1-1(d., x))
   g.1-0(e., g.1-1(e., x)) -> g.1-0(d., g.1-1(c., x))
   g.1-1(x, x) -> g.0-0(a., b.)

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

(31) 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:

   g.1-1(x, x) -> g.0-0(a., b.)

Used ordering: Polynomial interpretation [POLO]:

   POL(G.1-0(x_1, x_2)) = x_1 + x_2
   POL(a.) = 0
   POL(b.) = 0
   POL(c.) = 1
   POL(d.) = 1
   POL(e.) = 1
   POL(g.0-0(x_1, x_2)) = x_1 + x_2
   POL(g.1-0(x_1, x_2)) = x_1 + x_2
   POL(g.1-1(x_1, x_2)) = 1 + x_1 + x_2


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

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

   G.1-0(d., g.1-0(d., g.1-0(e., x0))) -> G.1-0(c., g.1-0(d., g.1-0(c., x0)))
   G.1-0(c., g.1-0(c., g.1-0(d., x0))) -> G.1-0(e., g.1-0(c., g.1-0(e., x0)))
   G.1-0(e., g.1-0(e., g.1-0(c., x0))) -> G.1-0(d., g.1-0(e., g.1-0(d., x0)))

The TRS R consists of the following rules:

   g.1-0(d., g.1-0(d., x)) -> g.1-0(c., g.1-0(e., x))
   g.1-0(c., g.1-0(c., x)) -> g.1-0(e., g.1-0(d., x))
   g.1-0(e., g.1-0(e., x)) -> g.1-0(d., g.1-0(c., x))
   g.1-0(d., g.1-1(d., x)) -> g.1-0(c., g.1-1(e., x))
   g.1-0(c., g.1-1(c., x)) -> g.1-0(e., g.1-1(d., x))
   g.1-0(e., g.1-1(e., x)) -> g.1-0(d., g.1-1(c., x))

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

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

(34)
TRUE

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

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

   G(d, g(d, g(e, x0))) -> G(c, g(d, g(c, x0)))
   G(c, g(c, g(d, x0))) -> G(e, g(c, g(e, x0)))
   G(e, g(e, g(c, x0))) -> G(d, g(e, g(d, x0)))

The TRS R consists of the following rules:

   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))
   g(e, g(e, x)) -> g(d, g(c, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(36) UsableRulesReductionPairsProof (EQUIVALENT)
First, we A-transformed [FROCOS05] the QDP-Problem.
Then we obtain the following A-transformed DP problem.

The pairs P are: 

   d1(d(e(x0))) -> c1(d(c(x0)))
   c1(c(d(x0))) -> e1(c(e(x0)))
   e1(e(c(x0))) -> d1(e(d(x0)))

and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:

   c(c(x)) -> e(d(x))
   e(e(x)) -> d(c(x))
   d(d(x)) -> c(e(x))

Q is empty.

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:

   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))
   g(e, g(e, x)) -> g(d, g(c, x))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(c(x_1)) = x_1
   POL(c1(x_1)) = 2*x_1
   POL(d(x_1)) = x_1
   POL(d1(x_1)) = 2*x_1
   POL(e(x_1)) = x_1
   POL(e1(x_1)) = 2*x_1


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

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

   d1(d(e(x0))) -> c1(d(c(x0)))
   c1(c(d(x0))) -> e1(c(e(x0)))
   e1(e(c(x0))) -> d1(e(d(x0)))

The TRS R consists of the following rules:

   c(c(x)) -> e(d(x))
   e(e(x)) -> d(c(x))
   d(d(x)) -> c(e(x))

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

(38) RFCMatchBoundsDPProof (EQUIVALENT)
Finiteness of the DP problem can be shown by a matchbound of 1. 
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

   d1(d(e(x0))) -> c1(d(c(x0)))
   c1(c(d(x0))) -> e1(c(e(x0)))
   e1(e(c(x0))) -> d1(e(d(x0)))

To find matches we regarded all rules of R and P:

   c(c(x)) -> e(d(x))
   e(e(x)) -> d(c(x))
   d(d(x)) -> c(e(x))
   d1(d(e(x0))) -> c1(d(c(x0)))
   c1(c(d(x0))) -> e1(c(e(x0)))
   e1(e(c(x0))) -> d1(e(d(x0)))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
703, 704, 705, 706, 707, 708, 709, 710, 711, 712, 713

Node 703 is start node and node 704 is final node.

Those nodes are connected through the following edges:

* 703 to 705 labelled c1_1(0)* 703 to 707 labelled e1_1(0)* 703 to 709 labelled d1_1(0)* 704 to 704 labelled #_1(0)* 705 to 706 labelled d_1(0)* 706 to 704 labelled c_1(0)* 706 to 711 labelled e_1(1)* 707 to 708 labelled c_1(0)* 708 to 704 labelled e_1(0)* 708 to 712 labelled d_1(1)* 709 to 710 labelled e_1(0)* 710 to 704 labelled d_1(0)* 710 to 713 labelled c_1(1)* 711 to 704 labelled d_1(1)* 711 to 713 labelled c_1(1)* 712 to 704 labelled c_1(1)* 712 to 711 labelled e_1(1)* 713 to 704 labelled e_1(1)* 713 to 712 labelled d_1(1)


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(39)
YES

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(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(g(x, y)) -> F(x)
   F(g(x, y)) -> F(f(x))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(c, g(c, x)) -> g(e, g(d, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(e, g(e, x)) -> g(d, g(c, x))
   f(g(x, y)) -> g(y, g(f(f(x)), a))

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

(41) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(g(x, y)) -> F(f(x)) at position [0] we obtained the following new rules [LPAR04]:

   (F(g(g(x0, x1), y1)) -> F(g(x1, g(f(f(x0)), a))),F(g(g(x0, x1), y1)) -> F(g(x1, g(f(f(x0)), a))))


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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(g(x, y)) -> F(x)
   F(g(g(x0, x1), y1)) -> F(g(x1, g(f(f(x0)), a)))

The TRS R consists of the following rules:

   g(x, x) -> g(a, b)
   g(c, g(c, x)) -> g(e, g(d, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(e, g(e, x)) -> g(d, g(c, x))
   f(g(x, y)) -> g(y, g(f(f(x)), a))

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

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


The following pairs can be oriented strictly and are deleted.

   F(g(x, y)) -> F(x)
   F(g(g(x0, x1), y1)) -> F(g(x1, g(f(f(x0)), a)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(F(x_1)) =  	[[0]] 	 +  	[[0, 2]] 	* 	x_1
>>>

   <<<
 POL(g(x_1, x_2)) =  	[[1], [1]] 	 +  	[[0, 0], [1, 1]] 	* 	x_1 	 +  	[[1, 1], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(f(x_1)) =  	[[0], [0]] 	 +  	[[0, 3], [1, 0]] 	* 	x_1
>>>

   <<<
 POL(a) =  	[[0], [0]]
>>>

   <<<
 POL(b) =  	[[0], [0]]
>>>

   <<<
 POL(e) =  	[[0], [0]]
>>>

   <<<
 POL(d) =  	[[0], [0]]
>>>

   <<<
 POL(c) =  	[[0], [0]]
>>>


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

   f(g(x, y)) -> g(y, g(f(f(x)), a))
   g(x, x) -> g(a, b)
   g(e, g(e, x)) -> g(d, g(c, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(c, g(c, x)) -> g(e, g(d, x))


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

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

   g(x, x) -> g(a, b)
   g(c, g(c, x)) -> g(e, g(d, x))
   g(d, g(d, x)) -> g(c, g(e, x))
   g(e, g(e, x)) -> g(d, g(c, x))
   f(g(x, y)) -> g(y, g(f(f(x)), a))

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

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

(46)
YES