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:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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:

MINUS₂(x, s₁(y)) -> P₁(minus₂(x, y))
FACTORIAL₁(x) -> FACITER₂(x, s₁(0))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(x, y))
FACITER₂(x, y) -> IF₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
IF₄(false, x, y, z) -> FACITER₂(x, z)
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
FACITER₂(x, y) -> MINUS₂(x, s₁(0))
FACITER₂(x, y) -> TIMES₂(y, x)
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
FACITER₂(x, y) -> ISZERO₁(x)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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:

MINUS₂(x, s₁(y)) -> P₁(minus₂(x, y))
FACTORIAL₁(x) -> FACITER₂(x, s₁(0))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(x, y))
FACITER₂(x, y) -> IF₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
IF₄(false, x, y, z) -> FACITER₂(x, z)
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
FACITER₂(x, y) -> MINUS₂(x, s₁(0))
FACITER₂(x, y) -> TIMES₂(y, x)
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
FACITER₂(x, y) -> ISZERO₁(x)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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.

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

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

POL(MINUS₂(x₁, x₂)) = 2·x₂
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:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

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

POL(PLUS₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 1 + 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:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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:

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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.

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

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

POL(TIMES₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 1 + 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:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(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:

FACITER₂(x, y) -> IF₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
IF₄(false, x, y, z) -> FACITER₂(x, z)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
p₁(s₁(x)) -> x
p₁(0) -> 0
minus₂(x, 0) -> x
minus₂(0, x) -> 0
minus₂(x, s₁(y)) -> p₁(minus₂(x, y))
isZero₁(0) -> true
isZero₁(s₁(x)) -> false
facIter₂(x, y) -> if₄(isZero₁(x), minus₂(x, s₁(0)), y, times₂(y, x))
if₄(true, x, y, z) -> y
if₄(false, x, y, z) -> facIter₂(x, z)
factorial₁(x) -> facIter₂(x, s₁(0))

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