Termination w.r.t. Q proof of /tmp/tmpU63PEQ/thiemann29.trs

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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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, b, y, z) → LOOP(x, b, times(b, y), s(z))
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
TIMES(s(x), y) → PLUS(y, times(x, y))
LOOP(x, s(s(b)), s(y), z) → LE(x, s(y))
TIMES(s(x), y) → TIMES(x, y)
LOG(s(x), s(s(b))) → LOOP(s(x), s(s(b)), s(0), 0)
IF(false, x, b, y, z) → TIMES(b, y)
LE(s(x), s(y)) → LE(x, y)
PLUS(s(x), y) → PLUS(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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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, b, y, z) → LOOP(x, b, times(b, y), s(z))
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
TIMES(s(x), y) → PLUS(y, times(x, y))
LOOP(x, s(s(b)), s(y), z) → LE(x, s(y))
TIMES(s(x), y) → TIMES(x, y)
LOG(s(x), s(s(b))) → LOOP(s(x), s(s(b)), s(0), 0)
IF(false, x, b, y, z) → TIMES(b, y)
LE(s(x), s(y)) → LE(x, y)
PLUS(s(x), y) → PLUS(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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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.

PLUS(s(x), y) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ 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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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

          ↳ QDP

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

TIMES(s(x), y) → TIMES(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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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.

TIMES(s(x), y) → TIMES(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(TIMES(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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

            ↳ 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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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)) = 1 + x_1
POL(LE(x₁, x₂)) = x_2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, 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

          ↳ QDP

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

IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), 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)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

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