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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) MNOCProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 33 ms]
(6) QDP
(7) PisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

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

(3) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

The set Q consists of the following terms:

   f(x0)

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

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


The following pairs can be oriented strictly and are deleted.

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

POL( F_1(x_1) ) = x_1 + 2
POL( f_1(x_1) ) = x_1 + 2
POL( s_1(x_1) ) = x_1 + 2

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

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))


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

(6)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

The set Q consists of the following terms:

   f(x0)

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

(7) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(8)
YES