YES Termination w.r.t. Q proof of AProVE_06_tower_sizeChange.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 70 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 AAECC Innermost (⇔, 0 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 0 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 AND
    ↳11 QDP
        ↳12 UsableRulesProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QReductionProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QReductionProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QReductionProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 TransformationProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 DependencyGraphProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 TransformationProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 TransformationProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 DependencyGraphProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 TransformationProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 SemLabProof (⇒, 58 ms)
        ↳50 QDP
        ↳51 DependencyGraphProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 UsableRulesReductionPairsProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 QDPOrderProof (⇔, 4 ms)
        ↳56 QDP
        ↳57 DependencyGraphProof (⇔, 0 ms)
        ↳58 TRUE

(0) Obligation:

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a) = 1   
POL(b) = 1   
POL(double(x1)) = x1   
POL(exp(x1)) = x1   
POL(f(x1, x2, x3)) = 2·x1 + 2·x2 + 2·x3   
POL(half(x1)) = x1   
POL(s(x1)) = x1   
POL(tower(x1)) = 2 + 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(a, 0, y) → y


(2) Obligation:

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

tower(x) → f(a, x, s(0))
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a) = 0   
POL(b) = 0   
POL(double(x1)) = x1   
POL(exp(x1)) = x1   
POL(f(x1, x2, x3)) = 2·x1 + 2·x2 + 2·x3   
POL(half(x1)) = x1   
POL(s(x1)) = x1   
POL(tower(x1)) = 2 + 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

tower(x) → f(a, x, s(0))


(4) Obligation:

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

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The TRS R 2 is

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))

The signature Sigma is {f}

(6) Obligation:

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

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → F(a, half(x), exp(y))
F(b, y, x) → HALF(x)
F(b, y, x) → EXP(y)
EXP(s(x)) → DOUBLE(exp(x))
EXP(s(x)) → EXP(x)
DOUBLE(s(x)) → DOUBLE(x)
HALF(0) → DOUBLE(0)
HALF(s(0)) → HALF(0)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(12) 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.

(13) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(14) 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(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(15) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

(16) 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:

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

(17) YES

(18) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(19) 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.

(20) Obligation:

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

HALF(s(s(x))) → HALF(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(21) 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(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(22) Obligation:

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

HALF(s(s(x))) → HALF(x)

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

(23) 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:

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

(24) YES

(25) Obligation:

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

EXP(s(x)) → EXP(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(26) 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.

(27) Obligation:

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

EXP(s(x)) → EXP(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(28) 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(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(29) Obligation:

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

EXP(s(x)) → EXP(x)

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

(30) 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:

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

(31) YES

(32) Obligation:

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(33) 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.

(34) Obligation:

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(35) 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(a, s(x0), x1)
f(b, x0, x1)

(36) Obligation:

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(37) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(b, y, x) → F(a, half(x), exp(y)) at position [1] we obtained the following new rules [LPAR04]:

F(b, y0, 0) → F(a, double(0), exp(y0)) → F(b, y0, 0) → F(a, double(0), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0)) → F(b, y0, s(0)) → F(a, half(0), exp(y0))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0)) → F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

(38) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, 0) → F(a, double(0), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(39) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(40) Obligation:

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

F(b, y0, s(0)) → F(a, half(0), exp(y0))
F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(41) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(b, y0, s(0)) → F(a, half(0), exp(y0)) at position [1] we obtained the following new rules [LPAR04]:

F(b, y0, s(0)) → F(a, double(0), exp(y0)) → F(b, y0, s(0)) → F(a, double(0), exp(y0))

(42) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, s(0)) → F(a, double(0), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(43) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(b, y0, s(0)) → F(a, double(0), exp(y0)) at position [1] we obtained the following new rules [LPAR04]:

F(b, y0, s(0)) → F(a, 0, exp(y0)) → F(b, y0, s(0)) → F(a, 0, exp(y0))

(44) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, s(0)) → F(a, 0, exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(45) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(46) Obligation:

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

F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(47) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(a, s(x), y) → F(b, y, s(x)) we obtained the following new rules [LPAR04]:

F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1))) → F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))

(48) Obligation:

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

F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(49) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1.
s: 0
exp: 0
a: 1
b: 0
half: 0
0: 0
F: 0
double: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.

(50) Obligation:

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

F.0-0-0(b., y0, s.0(s.0(x0))) → F.1-0-0(a., s.0(half.0(x0)), exp.0(y0))
F.1-0-0(a., s.0(s.0(y_1)), x1) → F.0-0-0(b., x1, s.0(s.0(y_1)))
F.1-0-1(a., s.0(s.0(y_1)), x1) → F.0-1-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(a., s.0(s.1(y_1)), x1) → F.0-0-0(b., x1, s.0(s.1(y_1)))
F.1-0-1(a., s.0(s.1(y_1)), x1) → F.0-1-0(b., x1, s.0(s.1(y_1)))
F.0-0-0(b., y0, s.0(s.1(x0))) → F.1-0-0(a., s.0(half.1(x0)), exp.0(y0))
F.0-1-0(b., y0, s.0(s.0(x0))) → F.1-0-0(a., s.0(half.0(x0)), exp.1(y0))
F.0-1-0(b., y0, s.0(s.1(x0))) → F.1-0-0(a., s.0(half.1(x0)), exp.1(y0))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
exp.0(s.1(x)) → double.0(exp.1(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
double.0(s.1(x)) → s.0(s.0(double.1(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(51) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.

(52) Obligation:

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

F.1-0-0(a., s.0(s.0(y_1)), x1) → F.0-0-0(b., x1, s.0(s.0(y_1)))
F.0-0-0(b., y0, s.0(s.0(x0))) → F.1-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
exp.0(s.1(x)) → double.0(exp.1(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
double.0(s.1(x)) → s.0(s.0(double.1(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(53) 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:

half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
double.0(s.1(x)) → s.0(s.0(double.1(x)))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0.) = 0   
POL(F.0-0-0(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(F.1-0-0(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(a.) = 0   
POL(b.) = 0   
POL(double.0(x1)) = x1   
POL(double.1(x1)) = x1   
POL(exp.0(x1)) = x1   
POL(exp.1(x1)) = x1   
POL(half.0(x1)) = x1   
POL(half.1(x1)) = x1   
POL(s.0(x1)) = x1   
POL(s.1(x1)) = x1   

(54) Obligation:

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

F.1-0-0(a., s.0(s.0(y_1)), x1) → F.0-0-0(b., x1, s.0(s.0(y_1)))
F.0-0-0(b., y0, s.0(s.0(x0))) → F.1-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(55) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


F.1-0-0(a., s.0(s.0(y_1)), x1) → F.0-0-0(b., x1, s.0(s.0(y_1)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0.) = 0   
POL(F.0-0-0(x1, x2, x3)) = x1 + x3   
POL(F.1-0-0(x1, x2, x3)) = 1 + x1 + x2   
POL(a.) = 1   
POL(b.) = 1   
POL(double.0(x1)) = 0   
POL(exp.0(x1)) = 1   
POL(half.0(x1)) = x1   
POL(s.0(x1)) = 1 + x1   

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

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.

(56) Obligation:

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

F.0-0-0(b., y0, s.0(s.0(x0))) → F.1-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(57) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(58) TRUE