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:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
f -> g
f -> h

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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:

PLUS2(x, y) -> EQ2(x, 0)
PLUS2(x, y) -> MINUS2(x, s1(0))
TIMES2(x, y) -> TIMESITER3(x, y, 0)
TIMESITER3(x, y, z) -> PLUS2(y, z)
TIMESITER3(x, y, z) -> IFTIMES5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
PLUS2(x, y) -> IFPLUS4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
IFTIMES5(false, x, y, z, u) -> TIMESITER3(x, y, u)
TIMESITER3(x, y, z) -> MINUS2(x, s1(0))
IFPLUS4(false, x, y, z) -> PLUS2(x, z)
EQ2(s1(x), s1(y)) -> EQ2(x, y)
TIMESITER3(x, y, z) -> EQ2(x, 0)
INC1(s1(x)) -> INC1(x)
PLUS2(x, y) -> INC1(x)

The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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:

PLUS2(x, y) -> EQ2(x, 0)
PLUS2(x, y) -> MINUS2(x, s1(0))
TIMES2(x, y) -> TIMESITER3(x, y, 0)
TIMESITER3(x, y, z) -> PLUS2(y, z)
TIMESITER3(x, y, z) -> IFTIMES5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
PLUS2(x, y) -> IFPLUS4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
IFTIMES5(false, x, y, z, u) -> TIMESITER3(x, y, u)
TIMESITER3(x, y, z) -> MINUS2(x, s1(0))
IFPLUS4(false, x, y, z) -> PLUS2(x, z)
EQ2(s1(x), s1(y)) -> EQ2(x, y)
TIMESITER3(x, y, z) -> EQ2(x, 0)
INC1(s1(x)) -> INC1(x)
PLUS2(x, y) -> INC1(x)

The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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 5 SCCs with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

EQ2(s1(x), s1(y)) -> EQ2(x, y)

The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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.


EQ2(s1(x), s1(y)) -> EQ2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( EQ2(x1, x2) ) = max{0, x2 - 1}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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
          ↳ QDP

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

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = max{0, x2 - 1}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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
          ↳ 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:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
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
          ↳ QDP

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

IFPLUS4(false, x, y, z) -> PLUS2(x, z)
PLUS2(x, y) -> IFPLUS4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))

The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
f -> g
f -> h

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TIMESITER3(x, y, z) -> IFTIMES5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
IFTIMES5(false, x, y, z, u) -> TIMESITER3(x, y, u)

The TRS R consists of the following rules:

inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
plus2(x, y) -> ifPlus4(eq2(x, 0), minus2(x, s1(0)), x, inc1(x))
ifPlus4(false, x, y, z) -> plus2(x, z)
ifPlus4(true, x, y, z) -> y
minus2(s1(x), s1(y)) -> minus2(x, y)
minus2(0, x) -> 0
minus2(x, 0) -> x
minus2(x, x) -> 0
eq2(s1(x), s1(y)) -> eq2(x, y)
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(0, 0) -> true
eq2(x, x) -> true
times2(x, y) -> timesIter3(x, y, 0)
timesIter3(x, y, z) -> ifTimes5(eq2(x, 0), minus2(x, s1(0)), y, z, plus2(y, z))
ifTimes5(true, x, y, z, u) -> z
ifTimes5(false, x, y, z, u) -> timesIter3(x, y, u)
f -> g
f -> h

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