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:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

Q is empty.

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

IF₄(true, true, x, y) -> LOGITER₂(x, y)
LOGITER₂(x, y) -> LE₂(s₁(0), x)
LOGARITHM₁(x) -> LOGITER₂(x, 0)
LOGITER₂(x, y) -> IF₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
LOGITER₂(x, y) -> LE₂(s₁(s₁(0)), x)
LOGITER₂(x, y) -> HALF₁(x)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
LOGITER₂(x, y) -> INC₁(y)
INC₁(s₁(x)) -> INC₁(x)
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

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:

IF₄(true, true, x, y) -> LOGITER₂(x, y)
LOGITER₂(x, y) -> LE₂(s₁(0), x)
LOGARITHM₁(x) -> LOGITER₂(x, 0)
LOGITER₂(x, y) -> IF₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
LOGITER₂(x, y) -> LE₂(s₁(s₁(0)), x)
LOGITER₂(x, y) -> HALF₁(x)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
LOGITER₂(x, y) -> INC₁(y)
INC₁(s₁(x)) -> INC₁(x)
HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INC₁(s₁(x)) -> INC₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

Q is empty.
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.

INC₁(s₁(x)) -> INC₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(INC₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] 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:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

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:

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

Q is empty.
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.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] 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:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

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:

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

Q is empty.
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.

HALF₁(s₁(s₁(x))) -> HALF₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HALF₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] 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:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

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:

IF₄(true, true, x, y) -> LOGITER₂(x, y)
LOGITER₂(x, y) -> IF₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))

The TRS R consists of the following rules:

half₁(0) -> 0
half₁(s₁(0)) -> 0
half₁(s₁(s₁(x))) -> s₁(half₁(x))
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
logarithm₁(x) -> logIter₂(x, 0)
logIter₂(x, y) -> if₄(le₂(s₁(0), x), le₂(s₁(s₁(0)), x), half₁(x), inc₁(y))
if₄(false, b, x, y) -> logZeroError
if₄(true, false, x, s₁(y)) -> y
if₄(true, true, x, y) -> logIter₂(x, y)
f -> g
f -> h

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