Termination w.r.t. Q proof of /tmp/tmpC0IhWo/log.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔)

        ↳9 QDP

        ↳10 QReductionProof (⇔)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔)

        ↳16 QDP

        ↳17 QReductionProof (⇔)

        ↳18 QDP

        ↳19 Narrowing (⇔)

        ↳20 QDP

        ↳21 DependencyGraphProof (⇔)

        ↳22 QDP

        ↳23 Narrowing (⇔)

        ↳24 QDP

        ↳25 DependencyGraphProof (⇔)

        ↳26 QDP

        ↳27 Narrowing (⇔)

        ↳28 QDP

        ↳29 DependencyGraphProof (⇔)

        ↳30 QDP

        ↳31 UsableRulesProof (⇔)

        ↳32 QDP

        ↳33 QReductionProof (⇔)

        ↳34 QDP

        ↳35 ForwardInstantiation (⇔)

        ↳36 QDP

        ↳37 ForwardInstantiation (⇔)

        ↳38 QDP

        ↳39 Rewriting (⇔)

        ↳40 QDP

        ↳41 UsableRulesProof (⇔)

        ↳42 QDP

        ↳43 Rewriting (⇔)

        ↳44 QDP

        ↳45 UsableRulesProof (⇔)

        ↳46 QDP

        ↳47 QReductionProof (⇔)

        ↳48 QDP

        ↳49 QDPSizeChangeProof (⇔)

        ↳50 YES

    ↳51 QDP

        ↳52 UsableRulesProof (⇔)

        ↳53 QDP

        ↳54 QReductionProof (⇔)

        ↳55 QDP

        ↳56 Rewriting (⇔)

        ↳57 QDP

        ↳58 Rewriting (⇔)

        ↳59 QDP

        ↳60 Narrowing (⇔)

        ↳61 QDP

        ↳62 DependencyGraphProof (⇔)

        ↳63 QDP

        ↳64 Rewriting (⇔)

        ↳65 QDP

        ↳66 Rewriting (⇔)

        ↳67 QDP

        ↳68 Rewriting (⇔)

        ↳69 QDP

        ↳70 Rewriting (⇔)

        ↳71 QDP

        ↳72 Rewriting (⇔)

        ↳73 QDP

        ↳74 Rewriting (⇔)

        ↳75 QDP

        ↳76 Rewriting (⇔)

        ↳77 QDP

            ↳78 MNOCProof (⇔)

                ↳79 QDP

            ↳80 MNOCProof (⇔)

                ↳81 QDP

(0) Obligation:

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

half(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

half(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

HALF(x) → IF(ge(x, s(s(0))), x)
HALF(x) → GE(x, s(s(0)))
IF(true, x) → HALF(p(p(x)))
IF(true, x) → P(p(x))
IF(true, x) → P(x)
GE(s(x), s(y)) → GE(x, y)
LOG(s(x)) → LOG(half(s(x)))
LOG(s(x)) → HALF(s(x))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(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:

GE(s(x), s(y)) → GE(x, y)

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

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(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].

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

(11) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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:

GE(s(x), s(y)) → GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

IF(true, x) → HALF(p(p(x)))
HALF(x) → IF(ge(x, s(s(0))), x)

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

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

(15) 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.

(16) Obligation:

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

IF(true, x) → HALF(p(p(x)))
HALF(x) → IF(ge(x, s(s(0))), x)

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

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

(17) 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].

half(x0)
if(false, x0)
if(true, x0)
log(0)
log(s(x0))

(18) Obligation:

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

IF(true, x) → HALF(p(p(x)))
HALF(x) → IF(ge(x, s(s(0))), x)

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule HALF(x) → IF(ge(x, s(s(0))), x) at position [0] we obtained the following new rules [LPAR04]:

HALF(0) → IF(false, 0)
HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))

(20) Obligation:

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

IF(true, x) → HALF(p(p(x)))
HALF(0) → IF(false, 0)
HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))
IF(true, x) → HALF(p(p(x)))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(23) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(true, x) → HALF(p(p(x))) at position [0] we obtained the following new rules [LPAR04]:

IF(true, 0) → HALF(p(0))
IF(true, s(x0)) → HALF(p(x0))

(24) Obligation:

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

HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))
IF(true, 0) → HALF(p(0))
IF(true, s(x0)) → HALF(p(x0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) Obligation:

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

IF(true, s(x0)) → HALF(p(x0))
HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(27) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(true, s(x0)) → HALF(p(x0)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(0)) → HALF(0)
IF(true, s(s(x0))) → HALF(x0)

(28) Obligation:

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

HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))
IF(true, s(0)) → HALF(0)
IF(true, s(s(x0))) → HALF(x0)

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) Obligation:

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

IF(true, s(s(x0))) → HALF(x0)
HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(31) 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.

(32) Obligation:

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

IF(true, s(s(x0))) → HALF(x0)
HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true

The set Q consists of the following terms:

p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(33) 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)
p(s(x0))

(34) Obligation:

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

IF(true, s(s(x0))) → HALF(x0)
HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(35) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(true, s(s(x0))) → HALF(x0) we obtained the following new rules [LPAR04]:

IF(true, s(s(s(y_0)))) → HALF(s(y_0))

(36) Obligation:

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

HALF(s(x0)) → IF(ge(x0, s(0)), s(x0))
IF(true, s(s(s(y_0)))) → HALF(s(y_0))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(37) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule HALF(s(x0)) → IF(ge(x0, s(0)), s(x0)) we obtained the following new rules [LPAR04]:

HALF(s(s(s(y_1)))) → IF(ge(s(s(y_1)), s(0)), s(s(s(y_1))))

(38) Obligation:

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

IF(true, s(s(s(y_0)))) → HALF(s(y_0))
HALF(s(s(s(y_1)))) → IF(ge(s(s(y_1)), s(0)), s(s(s(y_1))))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(39) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule HALF(s(s(s(y_1)))) → IF(ge(s(s(y_1)), s(0)), s(s(s(y_1)))) at position [0] we obtained the following new rules [LPAR04]:

HALF(s(s(s(y_1)))) → IF(ge(s(y_1), 0), s(s(s(y_1))))

(40) Obligation:

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

IF(true, s(s(s(y_0)))) → HALF(s(y_0))
HALF(s(s(s(y_1)))) → IF(ge(s(y_1), 0), s(s(s(y_1))))

The TRS R consists of the following rules:

ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(41) 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.

(42) Obligation:

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

IF(true, s(s(s(y_0)))) → HALF(s(y_0))
HALF(s(s(s(y_1)))) → IF(ge(s(y_1), 0), s(s(s(y_1))))

The TRS R consists of the following rules:

ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(43) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule HALF(s(s(s(y_1)))) → IF(ge(s(y_1), 0), s(s(s(y_1)))) at position [0] we obtained the following new rules [LPAR04]:

HALF(s(s(s(y_1)))) → IF(true, s(s(s(y_1))))

(44) Obligation:

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

IF(true, s(s(s(y_0)))) → HALF(s(y_0))
HALF(s(s(s(y_1)))) → IF(true, s(s(s(y_1))))

The TRS R consists of the following rules:

ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(45) 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.

(46) Obligation:

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

IF(true, s(s(s(y_0)))) → HALF(s(y_0))
HALF(s(s(s(y_1)))) → IF(true, s(s(s(y_1))))

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

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(47) 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].

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(48) Obligation:

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

IF(true, s(s(s(y_0)))) → HALF(s(y_0))
HALF(s(s(s(y_1)))) → IF(true, s(s(s(y_1))))

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

(49) 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(s(y_1)))) → IF(true, s(s(s(y_1))))
The graph contains the following edges 1 >= 2
IF(true, s(s(s(y_0)))) → HALF(s(y_0))
The graph contains the following edges 2 > 1

(50) YES

(51) Obligation:

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

LOG(s(x)) → LOG(half(s(x)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

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

(52) 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.

(53) Obligation:

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

LOG(s(x)) → LOG(half(s(x)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
log(0)
log(s(x0))

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

(54) 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].

log(0)
log(s(x0))

(55) Obligation:

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

LOG(s(x)) → LOG(half(s(x)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(56) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(x)) → LOG(half(s(x))) at position [0] we obtained the following new rules [LPAR04]:

LOG(s(x)) → LOG(if(ge(s(x), s(s(0))), s(x)))

(57) Obligation:

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

LOG(s(x)) → LOG(if(ge(s(x), s(s(0))), s(x)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(58) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(x)) → LOG(if(ge(s(x), s(s(0))), s(x))) at position [0,0] we obtained the following new rules [LPAR04]:

LOG(s(x)) → LOG(if(ge(x, s(0)), s(x)))

(59) Obligation:

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

LOG(s(x)) → LOG(if(ge(x, s(0)), s(x)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(60) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOG(s(x)) → LOG(if(ge(x, s(0)), s(x))) at position [0] we obtained the following new rules [LPAR04]:

LOG(s(0)) → LOG(if(false, s(0)))
LOG(s(s(x0))) → LOG(if(ge(x0, 0), s(s(x0))))

(61) Obligation:

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

LOG(s(0)) → LOG(if(false, s(0)))
LOG(s(s(x0))) → LOG(if(ge(x0, 0), s(s(x0))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(62) DependencyGraphProof (EQUIVALENT transformation)

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

(63) Obligation:

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

LOG(s(s(x0))) → LOG(if(ge(x0, 0), s(s(x0))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(64) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(if(ge(x0, 0), s(s(x0)))) at position [0,0] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(if(true, s(s(x0))))

(65) Obligation:

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

LOG(s(s(x0))) → LOG(if(true, s(s(x0))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(66) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(if(true, s(s(x0)))) at position [0] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(s(half(p(p(s(s(x0)))))))

(67) Obligation:

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

LOG(s(s(x0))) → LOG(s(half(p(p(s(s(x0)))))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(68) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(s(half(p(p(s(s(x0))))))) at position [0,0] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(s(if(ge(p(p(s(s(x0)))), s(s(0))), p(p(s(s(x0)))))))

(69) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(p(p(s(s(x0)))), s(s(0))), p(p(s(s(x0)))))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(70) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(s(if(ge(p(p(s(s(x0)))), s(s(0))), p(p(s(s(x0))))))) at position [0,0,0,0,0] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(s(if(ge(p(s(x0)), s(s(0))), p(p(s(s(x0)))))))

(71) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(p(s(x0)), s(s(0))), p(p(s(s(x0)))))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(72) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(s(if(ge(p(s(x0)), s(s(0))), p(p(s(s(x0))))))) at position [0,0,0,0] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), p(p(s(s(x0)))))))

(73) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), p(p(s(s(x0)))))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(74) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), p(p(s(s(x0))))))) at position [0,0,1,0] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), p(s(x0)))))

(75) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), p(s(x0)))))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(76) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), p(s(x0))))) at position [0,0,1] we obtained the following new rules [LPAR04]:

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), x0)))

(77) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), x0)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

The set Q consists of the following terms:

half(x0)
if(false, x0)
if(true, x0)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(78) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(79) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), x0)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

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

(80) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(81) Obligation:

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

LOG(s(s(x0))) → LOG(s(if(ge(x0, s(s(0))), x0)))

The TRS R consists of the following rules:

half(x) → if(ge(x, s(s(0))), x)
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true

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