Termination w.r.t. Q proof of TRS/Thiemannfactorial2.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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

      ↳ EdgeDeletionProof

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

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

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 deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

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 [13,14,18] contains 4 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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 [13].

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ 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:

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

      ↳ EdgeDeletionProof

        ↳ 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 [13].

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

Lexicographic Path Order [19].
Precedence:

s₁ > PLUS₁

The following usable rules [14] were oriented: none

↳ 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:

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

      ↳ EdgeDeletionProof

        ↳ 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 [13].

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

Lexicographic Path Order [19].
Precedence:

s₁ > TIMES₁

The following usable rules [14] were oriented: none

↳ 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:

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

      ↳ EdgeDeletionProof

        ↳ 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.