Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


DOUBLE(s(x)) → DOUBLE(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(DOUBLE(x1)) = (4)x_1   
POL(s(x1)) = 1 + (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 4 + (2)x_1   
POL(LE(x1, x2)) = (3)x_2   
The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

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