YES Termination w.r.t. Q proof of AProVE_07_kabasci06.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 ATransformationProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 MNOCProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 ATransformationProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 TRUE
    ↳23 QDP
        ↳24 QDPApplicativeOrderProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 PisEmptyProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QDPSizeChangeProof (⇔, 0 ms)
        ↳32 YES

(0) Obligation:

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

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, 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:

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, app(p, app(s, x)))
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, x))
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, y))
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(s, y))
APP(app(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), app(id, y)))
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), app(id, y))
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(div, app(s, x)), app(s, y)) → APP(id, y)
APP(id, x) → APP(s, app(s, app(s, x)))
APP(id, x) → APP(s, app(s, x))
APP(id, x) → APP(s, x)
APP(id, app(p, x)) → APP(id, app(s, app(id, x)))
APP(id, app(p, x)) → APP(s, app(id, x))
APP(id, app(p, x)) → APP(id, x)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, 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:

APP(id, app(p, x)) → APP(id, x)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, 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:

APP(id, app(p, x)) → APP(id, x)

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

(8) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(9) Obligation:

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

id(p(x)) → id(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:

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

(11) YES

(12) Obligation:

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, 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:

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x

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

(15) MNOCProof (EQUIVALENT transformation)

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

(16) Obligation:

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x

The set Q consists of the following terms:

app(p, app(s, x0))

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

(17) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(18) 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(x)) → x

The set Q consists of the following terms:

p(s(x0))

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

(19) 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:

p(s(x)) → x

Used ordering: Polynomial interpretation [POLO]:

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

(20) Obligation:

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

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

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

p(s(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

(23) Obligation:

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

APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(s, y))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

(24) QDPApplicativeOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].Here, we combined the reduction pair processor [LPAR04,JAR06] with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

div1(s(x), s(y)) → div1(minus(x, id(y)), s(y))

The a-transformed usable rules are

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


The following pairs can be oriented strictly and are deleted.


APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(s, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( p(x1) ) = max{0, x1 - 2}

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

POL( id(x1) ) = x1 + 2

POL( minus(x1, x2) ) = x1 + 1

POL( 0 ) = 0

POL( div1(x1, x2) ) = max{0, x1 + x2 - 2}


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

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

(25) Obligation:

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

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) YES

(28) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

(29) 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.

(30) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)

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

(31) 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:

  • APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
    The graph contains the following edges 1 >= 1, 2 > 2

  • APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
    The graph contains the following edges 1 > 1, 2 > 2

(32) YES