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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 46 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 2 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 107 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 1 ms]
(12) QDP
(13) PisEmptyProof [EQUIVALENT, 0 ms]
(14) YES


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

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

   a(s(x1)) -> s(s(s(p(s(b(p(p(s(s(x1))))))))))
   b(s(x1)) -> s(s(s(p(p(s(s(c(p(s(p(s(x1))))))))))))
   c(s(x1)) -> p(s(p(s(a(p(s(p(s(x1)))))))))
   p(p(s(x1))) -> p(x1)
   p(s(x1)) -> x1

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1

Q is empty.

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

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

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

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1

The set Q consists of the following terms:

   s(a(x0))
   s(b(x0))
   s(c(x0))
   s(p(x0))


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

(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:

   S(a(x1)) -> S(s(p(p(b(s(p(s(s(s(x1))))))))))
   S(a(x1)) -> S(p(p(b(s(p(s(s(s(x1)))))))))
   S(a(x1)) -> S(p(s(s(s(x1)))))
   S(a(x1)) -> S(s(s(x1)))
   S(a(x1)) -> S(s(x1))
   S(a(x1)) -> S(x1)
   S(b(x1)) -> S(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   S(b(x1)) -> S(p(c(s(s(p(p(s(s(s(x1))))))))))
   S(b(x1)) -> S(s(p(p(s(s(s(x1)))))))
   S(b(x1)) -> S(p(p(s(s(s(x1))))))
   S(b(x1)) -> S(s(s(x1)))
   S(b(x1)) -> S(s(x1))
   S(b(x1)) -> S(x1)
   S(c(x1)) -> S(p(s(p(a(s(p(s(p(x1)))))))))
   S(c(x1)) -> S(p(a(s(p(s(p(x1)))))))
   S(c(x1)) -> S(p(s(p(x1))))
   S(c(x1)) -> S(p(x1))

The TRS R consists of the following rules:

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1

The set Q consists of the following terms:

   s(a(x0))
   s(b(x0))
   s(c(x0))
   s(p(x0))

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 9 less nodes.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(a(x1)) -> S(s(s(x1)))
   S(a(x1)) -> S(s(p(p(b(s(p(s(s(s(x1))))))))))
   S(a(x1)) -> S(s(x1))
   S(a(x1)) -> S(x1)
   S(b(x1)) -> S(s(p(p(s(s(s(x1)))))))
   S(b(x1)) -> S(s(s(x1)))
   S(b(x1)) -> S(s(x1))
   S(b(x1)) -> S(x1)

The TRS R consists of the following rules:

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1

The set Q consists of the following terms:

   s(a(x0))
   s(b(x0))
   s(c(x0))
   s(p(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   S(a(x1)) -> S(s(s(x1)))
   S(a(x1)) -> S(s(x1))
   S(a(x1)) -> S(x1)
   S(b(x1)) -> S(s(p(p(s(s(s(x1)))))))
   S(b(x1)) -> S(s(s(x1)))
   S(b(x1)) -> S(s(x1))
   S(b(x1)) -> S(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( S_1(x_1) ) = max{0, 2x_1 - 1}
POL( s_1(x_1) ) = x_1
POL( a_1(x_1) ) = 2x_1 + 1
POL( p_1(x_1) ) = x_1
POL( b_1(x_1) ) = 2x_1 + 1
POL( c_1(x_1) ) = 2x_1 + 1

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

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1


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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(a(x1)) -> S(s(p(p(b(s(p(s(s(s(x1))))))))))

The TRS R consists of the following rules:

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1

The set Q consists of the following terms:

   s(a(x0))
   s(b(x0))
   s(c(x0))
   s(p(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   S(a(x1)) -> S(s(p(p(b(s(p(s(s(s(x1))))))))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(S(x_1)) = x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = 0
   POL(c(x_1)) = 1 + x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = x_1

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

   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1


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

(12)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   s(a(x1)) -> s(s(p(p(b(s(p(s(s(s(x1))))))))))
   s(b(x1)) -> s(p(s(p(c(s(s(p(p(s(s(s(x1))))))))))))
   s(c(x1)) -> s(p(s(p(a(s(p(s(p(x1)))))))))
   s(p(p(x1))) -> p(x1)
   s(p(x1)) -> x1

The set Q consists of the following terms:

   s(a(x0))
   s(b(x0))
   s(c(x0))
   s(p(x0))

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

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

(14)
YES