YES Termination w.r.t. Q proof of Secret_05_TRS_aprove5.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 11 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 MNOCProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 DependencyGraphProof (⇔, 0 ms)
        ↳16 AND
            ↳17 QDP
                ↳18 UsableRulesProof (⇔, 0 ms)
                ↳19 QDP
                ↳20 QReductionProof (⇔, 0 ms)
                ↳21 QDP
                ↳22 QDPOrderProof (⇔, 0 ms)
                ↳23 QDP
                ↳24 PisEmptyProof (⇔, 0 ms)
                ↳25 YES
            ↳26 QDP
                ↳27 QDPSizeChangeProof (⇔, 0 ms)
                ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 MNOCProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 UsableRulesReductionPairsProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 TransformationProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QDPOrderProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 PisEmptyProof (⇔, 0 ms)
        ↳43 YES
    ↳44 QDP
        ↳45 QDPOrderProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QDPOrderProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 PisEmptyProof (⇔, 0 ms)
        ↳50 YES

(0) Obligation:

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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:

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) → P(s(x))
MINUS(s(x), s(y)) → P(s(y))
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) → MINUS(x, y)
P(s(s(x))) → P(s(x))
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(s(x), s(y)) → MINUS(x, y)
DIV(plus(x, y), z) → PLUS(div(x, z), div(y, z))
DIV(plus(x, y), z) → DIV(x, z)
DIV(plus(x, y), z) → DIV(y, z)
PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))
PLUS(s(x), y) → MINUS(s(x), s(0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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:

P(s(s(x))) → P(s(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:

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

(9) YES

(10) Obligation:

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

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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:

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))

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:

MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
MINUS(x, plus(y, z)) → MINUS(x, y)

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(x0, plus(x1, x2))
minus(s(x0), s(x1))

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Complex Obligation (AND)

(17) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(x0, plus(x1, x2))
minus(s(x0), s(x1))

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

(18) 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.

(19) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(x0, plus(x1, x2))
minus(s(x0), s(x1))

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

(20) 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].

minus(x0, 0)
minus(0, x0)
minus(x0, plus(x1, x2))
minus(s(x0), s(x1))

(21) Obligation:

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

MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))

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

(22) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


MINUS(s(x), s(y)) → MINUS(p(s(x)), p(s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( MINUS(x1, x2) ) = x1 + 2x2

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

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


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

p(s(s(x))) → s(p(s(x)))

(23) Obligation:

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

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(s(x0)))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) YES

(26) Obligation:

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

MINUS(x, plus(y, z)) → MINUS(x, y)
MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))

The set Q consists of the following terms:

p(s(s(x0)))
minus(x0, 0)
minus(0, x0)
minus(x0, plus(x1, x2))
minus(s(x0), s(x1))

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

(27) 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:

  • MINUS(x, plus(y, z)) → MINUS(x, y)
    The graph contains the following edges 1 >= 1, 2 > 2

  • MINUS(x, plus(y, z)) → MINUS(minus(x, y), z)
    The graph contains the following edges 2 > 2

(28) YES

(29) Obligation:

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

PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(30) 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.

(31) Obligation:

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

PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, 0) → x
minus(0, y) → 0
p(s(s(x))) → s(p(s(x)))

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

(32) MNOCProof (EQUIVALENT transformation)

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

(33) Obligation:

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

PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, 0) → x
minus(0, y) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

minus(x0, plus(x1, x2))
minus(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(34) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

minus(x, plus(y, z)) → minus(minus(x, y), z)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(PLUS(x1, x2)) = 2·x1 + 2·x2   
POL(minus(x1, x2)) = x1 + x2   
POL(p(x1)) = x1   
POL(plus(x1, x2)) = 2 + x1 + x2   
POL(s(x1)) = x1   

(35) Obligation:

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

PLUS(s(x), y) → PLUS(y, minus(s(x), s(0)))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0

The set Q consists of the following terms:

minus(x0, plus(x1, x2))
minus(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(36) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule PLUS(s(x), y) → PLUS(y, minus(s(x), s(0))) at position [1] we obtained the following new rules [LPAR04]:

PLUS(s(x), y) → PLUS(y, minus(p(s(x)), p(s(0)))) → PLUS(s(x), y) → PLUS(y, minus(p(s(x)), p(s(0))))

(37) Obligation:

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

PLUS(s(x), y) → PLUS(y, minus(p(s(x)), p(s(0))))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(s(x))) → s(p(s(x)))
minus(x, 0) → x
minus(0, y) → 0

The set Q consists of the following terms:

minus(x0, plus(x1, x2))
minus(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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

(38) 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.

(39) Obligation:

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

PLUS(s(x), y) → PLUS(y, minus(p(s(x)), p(s(0))))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))
minus(0, y) → 0

The set Q consists of the following terms:

minus(x0, plus(x1, x2))
minus(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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.


PLUS(s(x), y) → PLUS(y, minus(p(s(x)), p(s(0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1, x2)
s(x1)  =  s
minus(x1, x2)  =  minus
0  =  0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s=3
0=1
minus=2
PLUS_2=1

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

minus(0, y) → 0

(41) Obligation:

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

p(s(s(x))) → s(p(s(x)))
minus(0, y) → 0

The set Q consists of the following terms:

minus(x0, plus(x1, x2))
minus(s(x0), s(x1))
minus(x0, 0)
minus(0, x0)
p(s(s(x0)))

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:

DIV(plus(x, y), z) → DIV(x, z)
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(plus(x, y), z) → DIV(y, z)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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

(45) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


DIV(plus(x, y), z) → DIV(x, z)
DIV(plus(x, y), z) → DIV(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2)  =  x1
plus(x1, x2)  =  plus(x1, x2)
s(x1)  =  x1
minus(x1, x2)  =  x1
0  =  0
p(x1)  =  p

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

0=2
p=1
plus_2=1

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

minus(x, 0) → x
minus(0, y) → 0
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(s(x))) → s(p(s(x)))

(46) Obligation:

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

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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.


DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2)  =  x1
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0
p(x1)  =  x1

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
0=1

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

minus(x, 0) → x
minus(0, y) → 0
minus(x, plus(y, z)) → minus(minus(x, y), z)
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
p(s(s(x))) → s(p(s(x)))

(48) Obligation:

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

minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) → minus(minus(x, y), z)
p(s(s(x))) → s(p(s(x)))
p(0) → s(s(0))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(plus(x, y), z) → plus(div(x, z), div(y, z))
plus(0, y) → y
plus(s(x), y) → s(plus(y, minus(s(x), s(0))))

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