/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 26 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 3 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 10 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) MNOCProof [EQUIVALENT, 2 ms]
        (15) QDP
        (16) MRRProof [EQUIVALENT, 1 ms]
        (17) QDP
        (18) PisEmptyProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) MNOCProof [EQUIVALENT, 1 ms]
        (24) QDP
        (25) MRRProof [EQUIVALENT, 9 ms]
        (26) QDP
        (27) PisEmptyProof [EQUIVALENT, 0 ms]
        (28) YES


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

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

   R(x1) -> r(x1)
   r(p(x1)) -> p(p(r(P(x1))))
   r(r(x1)) -> x1
   r(P(P(x1))) -> P(P(r(x1)))
   p(P(x1)) -> x1
   P(p(x1)) -> x1
   r(R(x1)) -> x1
   R(r(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:

   R(x1) -> r(x1)
   p(r(x1)) -> P(r(p(p(x1))))
   r(r(x1)) -> x1
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1
   R(r(x1)) -> x1
   r(R(x1)) -> x1

Q is empty.

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

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

   POL(P(x_1)) = x_1
   POL(R(x_1)) = 1 + x_1
   POL(p(x_1)) = x_1
   POL(r(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   R(x1) -> r(x1)
   R(r(x1)) -> x1
   r(R(x1)) -> x1




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

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

   p(r(x1)) -> P(r(p(p(x1))))
   r(r(x1)) -> x1
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1

Q is empty.

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

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

   POL(P(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(r(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   r(r(x1)) -> x1




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

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   p(r(x1)) -> P(r(p(p(x1))))
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1

Q is empty.

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

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

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

   P^1(r(x1)) -> P^2(r(p(p(x1))))
   P^1(r(x1)) -> P^1(p(x1))
   P^1(r(x1)) -> P^1(x1)
   P^2(P(r(x1))) -> P^2(P(x1))
   P^2(P(r(x1))) -> P^2(x1)

The TRS R consists of the following rules:

   p(r(x1)) -> P(r(p(p(x1))))
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1

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

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(10)
Complex Obligation (AND)

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

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

   P^2(P(r(x1))) -> P^2(x1)
   P^2(P(r(x1))) -> P^2(P(x1))

The TRS R consists of the following rules:

   p(r(x1)) -> P(r(p(p(x1))))
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1

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

(12) 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.
----------------------------------------

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

   P^2(P(r(x1))) -> P^2(x1)
   P^2(P(r(x1))) -> P^2(P(x1))

The TRS R consists of the following rules:

   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1

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

(14) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

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

   P^2(P(r(x1))) -> P^2(x1)
   P^2(P(r(x1))) -> P^2(P(x1))

The TRS R consists of the following rules:

   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1

The set Q consists of the following terms:

   P(P(r(x0)))
   P(p(x0))

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

(16) 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:

   P^2(P(r(x1))) -> P^2(x1)
   P^2(P(r(x1))) -> P^2(P(x1))

Strictly oriented rules of the TRS R:

   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1

Used ordering: Polynomial interpretation [POLO]:

   POL(P(x_1)) = 2 + 2*x_1
   POL(P^2(x_1)) = 2*x_1
   POL(p(x_1)) = 3 + x_1
   POL(r(x_1)) = 2 + x_1


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

(17)
Obligation:
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

   P(P(r(x0)))
   P(p(x0))

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

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

(19)
YES

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(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P^1(r(x1)) -> P^1(x1)
   P^1(r(x1)) -> P^1(p(x1))

The TRS R consists of the following rules:

   p(r(x1)) -> P(r(p(p(x1))))
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1

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

(21) 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.
----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P^1(r(x1)) -> P^1(x1)
   P^1(r(x1)) -> P^1(p(x1))

The TRS R consists of the following rules:

   p(r(x1)) -> P(r(p(p(x1))))
   p(P(x1)) -> x1

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

(23) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

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

   P^1(r(x1)) -> P^1(x1)
   P^1(r(x1)) -> P^1(p(x1))

The TRS R consists of the following rules:

   p(r(x1)) -> P(r(p(p(x1))))
   p(P(x1)) -> x1

The set Q consists of the following terms:

   p(r(x0))
   p(P(x0))

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

(25) 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:

   P^1(r(x1)) -> P^1(x1)
   P^1(r(x1)) -> P^1(p(x1))


Used ordering: Polynomial interpretation [POLO]:

   POL(P(x_1)) = x_1
   POL(P^1(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(r(x_1)) = 1 + 2*x_1


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

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

   p(r(x1)) -> P(r(p(p(x1))))
   p(P(x1)) -> x1

The set Q consists of the following terms:

   p(r(x0))
   p(P(x0))

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.
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(28)
YES