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


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 25 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 8 ms]
        (18) QDP
        (19) DependencyGraphProof [EQUIVALENT, 0 ms]
        (20) TRUE


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

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

Q is empty.

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

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

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

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   F(s(x)) -> MINUS(s(x), g(f(x)))
   F(s(x)) -> G(f(x))
   F(s(x)) -> F(x)
   G(s(x)) -> MINUS(s(x), f(g(x)))
   G(s(x)) -> F(g(x))
   G(s(x)) -> G(x)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

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

(6)
Complex Obligation (AND)

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

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

   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   MINUS(s(x), s(y)) -> MINUS(x, y)

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

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))


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

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

   MINUS(s(x), s(y)) -> MINUS(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*MINUS(s(x), s(y)) -> MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(13)
YES

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

   F(s(x)) -> G(f(x))
   G(s(x)) -> F(g(x))
   F(s(x)) -> F(x)
   G(s(x)) -> G(x)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   F(s(x)) -> F(x)
   G(s(x)) -> G(x)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1)  =  F(x1)

s(x1)  =  s(x1)

G(x1)  =  G(x1)

f(x1)  =  f(x1)

g(x1)  =  g(x1)

0  =  0

minus(x1, x2)  =  minus(x1)


Recursive path order with status [RPO].
Quasi-Precedence: [F_1, G_1] > [s_1, f_1, g_1, minus_1]
0 > [s_1, f_1, g_1, minus_1]

Status: F_1: multiset status
s_1: [1]
G_1: multiset status
f_1: [1]
g_1: [1]
0: multiset status
minus_1: [1]


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

   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)


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

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

   F(s(x)) -> G(f(x))
   G(s(x)) -> F(g(x))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   G(s(x)) -> F(g(x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( G_1(x_1) ) = 2x_1 + 1
POL( F_1(x_1) ) = x_1 + 2
POL( minus_2(x_1, x_2) ) = x_1
POL( f_1(x_1) ) = x_1 + 1
POL( 0 ) = 0
POL( s_1(x_1) ) = 2x_1 + 1
POL( g_1(x_1) ) = x_1

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

   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)


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

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

   F(s(x)) -> G(f(x))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(s(x), f(g(x)))

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))

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

(19) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(20)
TRUE