Termination w.r.t. Q proof of TRS/secret06logarithm.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:

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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
half(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError

The TRS R 2 is

f → g
f → h

The signature Sigma is {f, g, h}

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(x, y) → LE(s(0), x)
LOGARITHM(x) → LOGITER(x, 0)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → HALF(x)
LE(s(x), s(y)) → LE(x, y)
LOGITER(x, y) → INC(y)
INC(s(x)) → INC(x)
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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(x, y) → LE(s(0), x)
LOGARITHM(x) → LOGITER(x, 0)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → HALF(x)
LE(s(x), s(y)) → LE(x, y)
LOGITER(x, y) → INC(y)
INC(s(x)) → INC(x)
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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

LOGARITHM(x) → LOGITER(x, 0)
LOGITER(x, y) → LE(s(0), x)
IF(true, true, x, y) → LOGITER(x, y)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → HALF(x)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOGITER(x, y) → INC(y)
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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

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: Combined order from the following AFS and order.
INC(x1) = INC(x1)
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

s₁ > INC₁

Status:

INC₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

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: Combined order from the following AFS and order.
LE(x1, x2) = LE(x1)
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

s₁ > LE₁

Status:

LE₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

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: Combined order from the following AFS and order.
HALF(x1) = x1
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

trivial

Status:

s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(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(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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