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 Order [17,21] with Interpretation:

POL( INC1(x1) ) = 2x1 + 3


POL( s1(x1) ) = 3x1 + 3



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 Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



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 Order [17,21] with Interpretation:

POL( HALF1(x1) ) = max{0, 3x1 - 1}


POL( s1(x1) ) = 2x1 + 3



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.