YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_06_SRS/aprove01.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 35 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 MNOCProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 MNOCProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 MRRProof (⇔, 8 ms)
        ↳27 QDP
        ↳28 QDPOrderProof (⇔, 18 ms)
        ↳29 QDP
        ↳30 PisEmptyProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 MNOCProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 UsableRulesProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 QReductionProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 QDPOrderProof (⇔, 39 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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(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)) → P(x)
P(p(s(x))) → P(x)
F(s(x)) → P(s(g(p(s(s(x))))))
F(s(x)) → G(p(s(s(x))))
F(s(x)) → P(s(s(x)))
G(s(x)) → P(p(s(s(s(j(s(p(s(p(s(x)))))))))))
G(s(x)) → P(s(s(s(j(s(p(s(p(s(x))))))))))
G(s(x)) → J(s(p(s(p(s(x))))))
G(s(x)) → P(s(p(s(x))))
G(s(x)) → P(s(x))
J(s(x)) → P(s(s(p(s(f(p(s(p(p(s(x)))))))))))
J(s(x)) → P(s(f(p(s(p(p(s(x))))))))
J(s(x)) → F(p(s(p(p(s(x))))))
J(s(x)) → P(s(p(p(s(x)))))
J(s(x)) → P(p(s(x)))
J(s(x)) → P(s(x))
HALF(0(x)) → HALF(p(s(p(s(x)))))
HALF(0(x)) → P(s(p(s(x))))
HALF(0(x)) → P(s(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)) → 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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(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 15 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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

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

(6) MNOCProof (EQUIVALENT transformation)

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

(7) 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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

The set Q consists of the following terms:

p(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

(8) 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.

(9) Obligation:

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

RD(0(x)) → RD(x)

R is empty.
The set Q consists of the following terms:

p(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

(10) 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].

p(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

(11) 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.

(12) 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

(13) YES

(14) 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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

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

(15) 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.

(16) 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.

(17) 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

(18) YES

(19) 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(p(s(p(s(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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

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

(20) MNOCProof (EQUIVALENT transformation)

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

(21) 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(p(s(p(s(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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

The set Q consists of the following terms:

p(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

(22) 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.

(23) 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(p(s(p(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(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

(24) 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].

f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

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

(26) 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 dependency pairs:

HALF(0(x)) → HALF(p(s(p(s(x)))))


Used ordering: Polynomial interpretation [POLO]:

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

(27) 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(0(x0))
p(s(x0))

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

(28) 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))))

(29) 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(0(x0))
p(s(x0))

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) YES

(32) Obligation:

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

F(s(x)) → G(p(s(s(x))))
G(s(x)) → J(s(p(s(p(s(x))))))
J(s(x)) → F(p(s(p(p(s(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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

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

(33) MNOCProof (EQUIVALENT transformation)

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

(34) Obligation:

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

F(s(x)) → G(p(s(s(x))))
G(s(x)) → J(s(p(s(p(s(x))))))
J(s(x)) → F(p(s(p(p(s(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)) → p(s(g(p(s(s(x))))))
g(s(x)) → p(p(s(s(s(j(s(p(s(p(s(x)))))))))))
j(s(x)) → p(s(s(p(s(f(p(s(p(p(s(x)))))))))))
half(0(x)) → 0(s(s(half(p(s(p(s(x))))))))
half(s(s(x))) → s(half(p(p(s(s(x))))))
rd(0(x)) → 0(s(0(0(0(0(s(0(rd(x)))))))))

The set Q consists of the following terms:

p(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

(35) 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.

(36) Obligation:

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

F(s(x)) → G(p(s(s(x))))
G(s(x)) → J(s(p(s(p(s(x))))))
J(s(x)) → F(p(s(p(p(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(0(x0))
p(s(x0))
f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

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

(37) 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].

f(s(x0))
g(s(x0))
j(s(x0))
half(0(x0))
half(s(s(x0)))
rd(0(x0))

(38) Obligation:

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

F(s(x)) → G(p(s(s(x))))
G(s(x)) → J(s(p(s(p(s(x))))))
J(s(x)) → F(p(s(p(p(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(0(x0))
p(s(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.


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

POL( G(x1) ) = 2x1

POL( J(x1) ) = 2x1

POL( s(x1) ) = x1 + 2

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

(40) Obligation:

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

F(s(x)) → G(p(s(s(x))))
G(s(x)) → J(s(p(s(p(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(0(x0))
p(s(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