Termination w.r.t. Q proof of /tmp/tmpFbUgOI/factorial2.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:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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:

TIMES(s(x), y) → PLUS(y, times(x, y))
FACITER(x, y) → IF(isZero(x), minus(x, s(0)), y, times(y, x))
FACITER(x, y) → TIMES(y, x)
MINUS(x, s(y)) → P(minus(x, y))
TIMES(s(x), y) → TIMES(x, y)
IF(false, x, y, z) → FACITER(x, z)
FACITER(x, y) → MINUS(x, s(0))
FACITER(x, y) → ISZERO(x)
MINUS(x, s(y)) → MINUS(x, y)
FACTORIAL(x) → FACITER(x, s(0))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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:

TIMES(s(x), y) → PLUS(y, times(x, y))
FACITER(x, y) → IF(isZero(x), minus(x, s(0)), y, times(y, x))
FACITER(x, y) → TIMES(y, x)
MINUS(x, s(y)) → P(minus(x, y))
TIMES(s(x), y) → TIMES(x, y)
IF(false, x, y, z) → FACITER(x, z)
FACITER(x, y) → MINUS(x, s(0))
FACITER(x, y) → ISZERO(x)
MINUS(x, s(y)) → MINUS(x, y)
FACTORIAL(x) → FACITER(x, s(0))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS(x, s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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.

MINUS(x, s(y)) → MINUS(x, y)

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

POL(MINUS(x₁, x₂)) = (1/4)x_2
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
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:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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.

PLUS(s(x), y) → PLUS(x, y)

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

POL(PLUS(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
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:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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:

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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.

TIMES(s(x), y) → TIMES(x, y)

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

POL(TIMES(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
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:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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:

FACITER(x, y) → IF(isZero(x), minus(x, s(0)), y, times(y, x))
IF(false, x, y, z) → FACITER(x, z)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
p(s(x)) → x
p(0) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(x, s(y)) → p(minus(x, y))
isZero(0) → true
isZero(s(x)) → false
facIter(x, y) → if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z) → y
if(false, x, y, z) → facIter(x, z)
factorial(x) → facIter(x, s(0))

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