YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_06_SRS/aprove02.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 30 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 2 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 1 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 MNOCProof (⇔, 9 ms)
        ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPOrderProof (⇔, 109 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

v(s(x)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))))
v(0(x)) → p(p(s(s(0(p(p(s(s(s(s(s(x))))))))))))
w(s(x)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))))))))
w(0(x)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x)))))))))))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   
POL(v(x1)) = 1 + x1   
POL(w(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

v(0(x)) → p(p(s(s(0(p(p(s(s(s(s(s(x))))))))))))
w(0(x)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x)))))))))))))))))))


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

v(s(x)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))))
w(s(x)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

V(s(x)) → P(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x)))))))))))))))))))
V(s(x)) → P(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))
V(s(x)) → W(p(p(s(s(p(s(p(s(x)))))))))
V(s(x)) → P(p(s(s(p(s(p(s(x))))))))
V(s(x)) → P(s(s(p(s(p(s(x)))))))
V(s(x)) → P(s(p(s(x))))
V(s(x)) → P(s(x))
W(s(x)) → P(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))
W(s(x)) → P(s(s(v(p(p(s(s(s(p(p(s(s(x)))))))))))))
W(s(x)) → V(p(p(s(s(s(p(p(s(s(x))))))))))
W(s(x)) → P(p(s(s(s(p(p(s(s(x)))))))))
W(s(x)) → P(s(s(s(p(p(s(s(x))))))))
W(s(x)) → P(p(s(s(x))))
W(s(x)) → P(s(s(x)))
P(p(s(x))) → P(x)
P(0(x)) → P(s(x))

The TRS R consists of the following rules:

v(s(x)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))))
w(s(x)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 13 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

P(p(s(x))) → P(x)

The TRS R consists of the following rules:

v(s(x)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))))
w(s(x)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

P(p(s(x))) → P(x)

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

(10) QDPSizeChangeProof (EQUIVALENT transformation)

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:

  • P(p(s(x))) → P(x)
    The graph contains the following edges 1 > 1

(11) YES

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

V(s(x)) → W(p(p(s(s(p(s(p(s(x)))))))))
W(s(x)) → V(p(p(s(s(s(p(p(s(s(x))))))))))

The TRS R consists of the following rules:

v(s(x)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))))
w(s(x)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

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

(13) MNOCProof (EQUIVALENT transformation)

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

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

V(s(x)) → W(p(p(s(s(p(s(p(s(x)))))))))
W(s(x)) → V(p(p(s(s(s(p(p(s(s(x))))))))))

The TRS R consists of the following rules:

v(s(x)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x))))))))))))))))))))
w(s(x)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x))))))))))))))))))))
p(p(s(x))) → p(x)
p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

The set Q consists of the following terms:

v(s(x0))
w(s(x0))
p(s(x0))
p(0(x0))

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

(15) UsableRulesProof (EQUIVALENT transformation)

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.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

V(s(x)) → W(p(p(s(s(p(s(p(s(x)))))))))
W(s(x)) → V(p(p(s(s(s(p(p(s(s(x))))))))))

The TRS R consists of the following rules:

p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

The set Q consists of the following terms:

v(s(x0))
w(s(x0))
p(s(x0))
p(0(x0))

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

(17) QReductionProof (EQUIVALENT transformation)

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

v(s(x0))
w(s(x0))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

V(s(x)) → W(p(p(s(s(p(s(p(s(x)))))))))
W(s(x)) → V(p(p(s(s(s(p(p(s(s(x))))))))))

The TRS R consists of the following rules:

p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


W(s(x)) → V(p(p(s(s(s(p(p(s(s(x))))))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( V(x1) ) = max{0, x1 - 2}

POL( W(x1) ) = x1

POL( p(x1) ) = max{0, x1 - 2}

POL( s(x1) ) = x1 + 2

POL( 0(x1) ) = 0


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

p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

V(s(x)) → W(p(p(s(s(p(s(p(s(x)))))))))

The TRS R consists of the following rules:

p(s(x)) → x
p(0(x)) → 0(s(s(s(s(s(s(s(p(s(x))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(22) TRUE