Termination w.r.t. Q proof of /tmp/tmpWAcsrf/thiemann12.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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:

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → HALF(x)
LOG2(x, y) → LE(x, s(0))
HALF(s(s(x))) → HALF(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
INC(s(x)) → INC(x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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:

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → HALF(x)
LOG2(x, y) → LE(x, s(0))
HALF(s(s(x))) → HALF(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
INC(s(x)) → INC(x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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:

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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.

INC(s(x)) → INC(x)

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

POL(INC(x₁)) = (2)x_1
POL(s(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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)) = 1/2 + (13/4)x_1
POL(LE(x₁, x₂)) = (15/4)x_2

The value of delta used in the strict ordering is 15/8.
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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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.

HALF(s(s(x))) → HALF(x)

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

POL(HALF(x₁)) = (1/2)x_1
POL(s(x₁)) = 1/4 + x_1

The value of delta used in the strict ordering is 1/4.
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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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

            ↳ QDPOrderProof

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

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

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.

IF(false, x, y) → LOG2(half(x), y)

The remaining pairs can at least be oriented weakly.

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

Used ordering: Polynomial interpretation [25,35]:

POL(LOG2(x₁, x₂)) = (4)x_1
POL(half(x₁)) = (1/2)x_1
POL(le(x₁, x₂)) = x_1
POL(true) = 0
POL(false) = 1/2
POL(s(x₁)) = 1/2 + (3/2)x_1
POL(inc(x₁)) = 4
POL(IF(x₁, x₂, x₃)) = (1/2)x_1 + (2)x_2
POL(0) = 0

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.