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

(0) Obligation:

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

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

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:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

(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:

LE(s(x), s(y)) → LE(x, y)
DOUBLE(s(x)) → DOUBLE(x)
LOG(s(x)) → LOOP(s(x), s(0), 0)
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
LOOP(x, s(y), z) → LE(x, s(y))
IF(false, x, y, z) → LOOP(x, double(y), s(z))
IF(false, x, y, z) → DOUBLE(y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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 3 less nodes.

(6) Complex Obligation (AND)

(7) 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:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DOUBLE(s(x)) → DOUBLE(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DOUBLE(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
IF(false, x, y, z) → LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

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