/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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, 8 ms]
(4) QTRS
(5) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) MRRProof [EQUIVALENT, 38 ms]
(12) QDP
(13) PisEmptyProof [EQUIVALENT, 0 ms]
(14) YES


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

   f(s(x1)) -> s(s(f(p(s(x1)))))
   f(0(x1)) -> 0(x1)
   p(s(x1)) -> x1

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   s(f(x1)) -> s(p(f(s(s(x1)))))
   0(f(x1)) -> 0(x1)
   s(p(x1)) -> x1

Q is empty.

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

   POL(0(x_1)) = x_1
   POL(f(x_1)) = 1 + x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   0(f(x1)) -> 0(x1)




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

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

Q is empty.

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(5) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   s(f(x0))
   s(p(x0))


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(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(f(x1)) -> S(p(f(s(s(x1)))))
   S(f(x1)) -> S(s(x1))
   S(f(x1)) -> S(x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   s(f(x0))
   s(p(x0))

We have to consider all minimal (P,Q,R)-chains.
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(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(f(x1)) -> S(x1)
   S(f(x1)) -> S(s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   s(f(x0))
   s(p(x0))

We have to consider all minimal (P,Q,R)-chains.
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(11) 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:

   S(f(x1)) -> S(x1)
   S(f(x1)) -> S(s(x1))


Used ordering: Polynomial interpretation [POLO]:

   POL(S(x_1)) = x_1
   POL(f(x_1)) = 1 + x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = x_1


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

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

The set Q consists of the following terms:

   s(f(x0))
   s(p(x0))

We have to consider all minimal (P,Q,R)-chains.
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(13) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(14)
YES