YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_05_SRS/aprove1.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 6 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPOrderProof (⇔, 5 ms)
        ↳19 QDP
        ↳20 MRRProof (⇔, 12 ms)
        ↳21 QDP
        ↳22 MNOCProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPOrderProof (⇔, 14 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 MRRProof (⇔, 26 ms)
        ↳34 QDP
        ↳35 MNOCProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 UsableRulesProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 QDPOrderProof (⇔, 159 ms)
        ↳40 QDP
        ↳41 DependencyGraphProof (⇔, 0 ms)
        ↳42 TRUE

(0) Obligation:

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

p(0(x)) → 0(s(s(p(x))))
p(s(x)) → x
p(p(s(x))) → p(x)
f(s(x)) → g(s(x))
g(x) → i(s(half(x)))
i(x) → f(p(x))
half(0(x)) → 0(s(s(half(x))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
0(x) → x
rd(0(x)) → 0(0(0(0(0(0(rd(x)))))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

P(0(x)) → 01(s(s(p(x))))
P(0(x)) → P(x)
P(p(s(x))) → P(x)
F(s(x)) → G(s(x))
G(x) → I(s(half(x)))
G(x) → HALF(x)
I(x) → F(p(x))
I(x) → P(x)
HALF(0(x)) → 01(s(s(half(x))))
HALF(0(x)) → HALF(x)
HALF(s(s(x))) → HALF(p(p(s(s(x)))))
HALF(s(s(x))) → P(p(s(s(x))))
HALF(s(s(x))) → P(s(s(x)))
RD(0(x)) → 01(0(0(0(0(0(rd(x)))))))
RD(0(x)) → 01(0(0(0(0(rd(x))))))
RD(0(x)) → 01(0(0(0(rd(x)))))
RD(0(x)) → 01(0(0(rd(x))))
RD(0(x)) → 01(0(rd(x)))
RD(0(x)) → 01(rd(x))
RD(0(x)) → RD(x)

The TRS R consists of the following rules:

p(0(x)) → 0(s(s(p(x))))
p(s(x)) → x
p(p(s(x))) → p(x)
f(s(x)) → g(s(x))
g(x) → i(s(half(x)))
i(x) → f(p(x))
half(0(x)) → 0(s(s(half(x))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
0(x) → x
rd(0(x)) → 0(0(0(0(0(0(rd(x)))))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 12 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

RD(0(x)) → RD(x)

The TRS R consists of the following rules:

p(0(x)) → 0(s(s(p(x))))
p(s(x)) → x
p(p(s(x))) → p(x)
f(s(x)) → g(s(x))
g(x) → i(s(half(x)))
i(x) → f(p(x))
half(0(x)) → 0(s(s(half(x))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
0(x) → x
rd(0(x)) → 0(0(0(0(0(0(rd(x)))))))

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

(6) 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.

(7) Obligation:

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

RD(0(x)) → RD(x)

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

(8) 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:

  • RD(0(x)) → RD(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

P(p(s(x))) → P(x)
P(0(x)) → P(x)

The TRS R consists of the following rules:

p(0(x)) → 0(s(s(p(x))))
p(s(x)) → x
p(p(s(x))) → p(x)
f(s(x)) → g(s(x))
g(x) → i(s(half(x)))
i(x) → f(p(x))
half(0(x)) → 0(s(s(half(x))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
0(x) → x
rd(0(x)) → 0(0(0(0(0(0(rd(x)))))))

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

(11) 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.

(12) Obligation:

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

P(p(s(x))) → P(x)
P(0(x)) → P(x)

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

(13) 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

  • P(0(x)) → P(x)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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

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

The TRS R consists of the following rules:

p(0(x)) → 0(s(s(p(x))))
p(s(x)) → x
p(p(s(x))) → p(x)
f(s(x)) → g(s(x))
g(x) → i(s(half(x)))
i(x) → f(p(x))
half(0(x)) → 0(s(s(half(x))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
0(x) → x
rd(0(x)) → 0(0(0(0(0(0(rd(x)))))))

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

(16) 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.

(17) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(18) 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(p(x1)) = x1   
POL(s(x1)) = x1   

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(p(x))))
p(p(s(x))) → p(x)
0(x) → x

(19) Obligation:

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

HALF(s(s(x))) → HALF(p(p(s(s(x)))))

The TRS R consists of the following rules:

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

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

(20) 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:

0(x) → x

Used ordering: Polynomial interpretation [POLO]:

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

(21) Obligation:

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

HALF(s(s(x))) → HALF(p(p(s(s(x)))))

The TRS R consists of the following rules:

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

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

(22) MNOCProof (EQUIVALENT transformation)

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

(23) Obligation:

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

HALF(s(s(x))) → HALF(p(p(s(s(x)))))

The TRS R consists of the following rules:

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

(24) 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.

(25) Obligation:

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

HALF(s(s(x))) → HALF(p(p(s(s(x)))))

The TRS R consists of the following rules:

p(s(x)) → x
p(0(x)) → 0(s(s(p(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.

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

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

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

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


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(p(x))))

(27) Obligation:

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

p(s(x)) → x
p(0(x)) → 0(s(s(p(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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES

(30) Obligation:

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

I(x) → F(p(x))
F(s(x)) → G(s(x))
G(x) → I(s(half(x)))

The TRS R consists of the following rules:

p(0(x)) → 0(s(s(p(x))))
p(s(x)) → x
p(p(s(x))) → p(x)
f(s(x)) → g(s(x))
g(x) → i(s(half(x)))
i(x) → f(p(x))
half(0(x)) → 0(s(s(half(x))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
0(x) → x
rd(0(x)) → 0(0(0(0(0(0(rd(x)))))))

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

(31) 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.

(32) Obligation:

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

I(x) → F(p(x))
F(s(x)) → G(s(x))
G(x) → I(s(half(x)))

The TRS R consists of the following rules:

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

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

(33) 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:

0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 2 + x1   
POL(F(x1)) = x1   
POL(G(x1)) = x1   
POL(I(x1)) = x1   
POL(half(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   

(34) Obligation:

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

I(x) → F(p(x))
F(s(x)) → G(s(x))
G(x) → I(s(half(x)))

The TRS R consists of the following rules:

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

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

(35) MNOCProof (EQUIVALENT transformation)

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

(36) Obligation:

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

I(x) → F(p(x))
F(s(x)) → G(s(x))
G(x) → I(s(half(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(37) 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.

(38) Obligation:

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

I(x) → F(p(x))
F(s(x)) → G(s(x))
G(x) → I(s(half(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(39) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


F(s(x)) → G(s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(I(x1)) = 0A +
[-I,0A,0A]
·x1

POL(F(x1)) = 0A +
[0A,0A,0A]
·x1

POL(p(x1)) =
/0A\
|-I|
\0A/
 +
/-I0A0A\
|-I0A-I|
\-I0A-I/
·x1

POL(s(x1)) =
/1A\
|0A|
\0A/
 +
/1A1A1A\
|-I0A0A|
\0A0A0A/
·x1

POL(G(x1)) = 0A +
[-I,0A,0A]
·x1

POL(half(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A0A\
|-I0A0A|
\-I0A0A/
·x1

POL(0(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1

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

(40) Obligation:

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

I(x) → F(p(x))
G(x) → I(s(half(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(41) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(42) TRUE