YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_05_SRS/aprove4.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 45 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 11 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES
    ↳17 QDP
        ↳18 UsableRulesProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPOrderProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 DependencyGraphProof (⇔, 0 ms)
        ↳23 AND
            ↳24 QDP
                ↳25 UsableRulesProof (⇔, 0 ms)
                ↳26 QDP
                ↳27 MRRProof (⇔, 4 ms)
                ↳28 QDP
                ↳29 MNOCProof (⇔, 0 ms)
                ↳30 QDP
                ↳31 QDPOrderProof (⇔, 9 ms)
                ↳32 QDP
                ↳33 PisEmptyProof (⇔, 0 ms)
                ↳34 YES
            ↳35 QDP
                ↳36 QDPOrderProof (⇔, 11 ms)
                ↳37 QDP
                ↳38 MRRProof (⇔, 8 ms)
                ↳39 QDP
                ↳40 QDPOrderProof (⇔, 15 ms)
                ↳41 QDP
                ↳42 PisEmptyProof (⇔, 0 ms)
                ↳43 YES
    ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QDPOrderProof (⇔, 88 ms)
        ↳48 QDP
        ↳49 PisEmptyProof (⇔, 0 ms)
        ↳50 YES

(0) Obligation:

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

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(s(x))
0(x) → x

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

0(x) → x


(2) Obligation:

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

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(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:

LOG(s(x)) → S(log(half(s(x))))
LOG(s(x)) → LOG(half(s(x)))
LOG(s(x)) → HALF(s(x))
HALF(0(x)) → S(s(half(x)))
HALF(0(x)) → S(half(x))
HALF(0(x)) → HALF(x)
HALF(s(s(x))) → S(half(p(s(s(x)))))
HALF(s(s(x))) → HALF(p(s(s(x))))
HALF(s(s(x))) → P(s(s(x)))
HALF(half(s(s(s(s(x)))))) → S(s(half(half(x))))
HALF(half(s(s(s(s(x)))))) → S(half(half(x)))
HALF(half(s(s(s(s(x)))))) → HALF(half(x))
HALF(half(s(s(s(s(x)))))) → HALF(x)
P(s(s(s(x)))) → S(p(s(s(x))))
P(s(s(s(x)))) → P(s(s(x)))
S(s(p(s(x)))) → S(s(x))

The TRS R consists of the following rules:

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(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 4 SCCs with 9 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

S(s(p(s(x)))) → S(s(x))

The TRS R consists of the following rules:

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(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:

S(s(p(s(x)))) → S(s(x))

The TRS R consists of the following rules:

s(s(p(s(x)))) → s(s(x))

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:

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

(11) YES

(12) Obligation:

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

P(s(s(s(x)))) → P(s(s(x)))

The TRS R consists of the following rules:

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(s(x))

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

(13) 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.

(14) Obligation:

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

P(s(s(s(x)))) → P(s(s(x)))

The TRS R consists of the following rules:

s(s(p(s(x)))) → s(s(x))

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

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

(16) YES

(17) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))
HALF(0(x)) → HALF(x)
HALF(half(s(s(s(s(x)))))) → HALF(half(x))
HALF(half(s(s(s(s(x)))))) → HALF(x)

The TRS R consists of the following rules:

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(s(x))

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

(18) 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.

(19) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))
HALF(0(x)) → HALF(x)
HALF(half(s(s(s(s(x)))))) → HALF(half(x))
HALF(half(s(s(s(s(x)))))) → HALF(x)

The TRS R consists of the following rules:

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

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

(20) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


HALF(0(x)) → HALF(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(HALF(x1)) = x1   
POL(half(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   

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

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

(21) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))
HALF(half(s(s(s(s(x)))))) → HALF(half(x))
HALF(half(s(s(s(s(x)))))) → HALF(x)

The TRS R consists of the following rules:

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

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

(22) DependencyGraphProof (EQUIVALENT transformation)

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

(23) Complex Obligation (AND)

(24) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))

The TRS R consists of the following rules:

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

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

(25) 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.

(26) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))

The TRS R consists of the following rules:

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

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

(27) MRRProof (EQUIVALENT transformation)

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 rules of the TRS R:

s(s(p(s(x)))) → s(s(x))

Used ordering: Polynomial interpretation [POLO]:

POL(HALF(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 2 + 2·x1   

(28) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))

The TRS R consists of the following rules:

p(s(s(s(x)))) → s(p(s(s(x))))

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

(29) MNOCProof (EQUIVALENT transformation)

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

(30) Obligation:

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

HALF(s(s(x))) → HALF(p(s(s(x))))

The TRS R consists of the following rules:

p(s(s(s(x)))) → s(p(s(s(x))))

The set Q consists of the following terms:

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

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

(31) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

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

POL( s(x1) ) = 2x1 + 1


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

p(s(s(s(x)))) → s(p(s(s(x))))

(32) Obligation:

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

p(s(s(s(x)))) → s(p(s(s(x))))

The set Q consists of the following terms:

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

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

(33) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(34) YES

(35) Obligation:

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

HALF(half(s(s(s(s(x)))))) → HALF(x)
HALF(half(s(s(s(s(x)))))) → HALF(half(x))

The TRS R consists of the following rules:

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

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

(36) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


HALF(half(s(s(s(s(x)))))) → HALF(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(HALF(x1)) = x1   
POL(half(x1)) = 1 + x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   

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

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

(37) Obligation:

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

HALF(half(s(s(s(s(x)))))) → HALF(half(x))

The TRS R consists of the following rules:

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

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

(38) MRRProof (EQUIVALENT transformation)

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 rules of the TRS R:

half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)

Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + 2·x1   
POL(HALF(x1)) = x1   
POL(half(x1)) = 2·x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   

(39) Obligation:

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

HALF(half(s(s(s(s(x)))))) → HALF(half(x))

The TRS R consists of the following rules:

half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
s(s(p(s(x)))) → s(s(x))
p(s(s(s(x)))) → s(p(s(s(x))))

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

(40) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

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

POL( s(x1) ) = x1 + 2

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


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

half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
s(s(p(s(x)))) → s(s(x))
p(s(s(s(x)))) → s(p(s(s(x))))

(41) Obligation:

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

half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
s(s(p(s(x)))) → s(s(x))
p(s(s(s(x)))) → s(p(s(s(x))))

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

(42) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(43) YES

(44) Obligation:

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

LOG(s(x)) → LOG(half(s(x)))

The TRS R consists of the following rules:

log(s(x)) → s(log(half(s(x))))
half(0(x)) → 0(s(s(half(x))))
half(s(0(x))) → 0(x)
half(s(s(x))) → s(half(p(s(s(x)))))
half(half(s(s(s(s(x)))))) → s(s(half(half(x))))
p(s(s(s(x)))) → s(p(s(s(x))))
s(s(p(s(x)))) → s(s(x))

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

(45) 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.

(46) Obligation:

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

LOG(s(x)) → LOG(half(s(x)))

The TRS R consists of the following rules:

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

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

(47) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


LOG(s(x)) → LOG(half(s(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( LOG(x1) ) = x1 + 2

POL( s(x1) ) = x1 + 1

POL( p(x1) ) = max{0, x1 - 1}

POL( half(x1) ) = max{0, x1 - 1}

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


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

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

(48) Obligation:

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

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

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

(49) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(50) YES