Termination w.r.t. Q proof of /tmp/tmpPQYAX9/perfect2.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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:

F(s(x), s(y), z, u) → MINUS(y, x)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), s(y), z, u) → IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x), s(y), z, u) → LE(x, y)
F(s(x), 0, z, u) → MINUS(z, s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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:

F(s(x), s(y), z, u) → MINUS(y, x)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), s(y), z, u) → IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x), s(y), z, u) → LE(x, y)
F(s(x), 0, z, u) → MINUS(z, s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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.

LE(s(x), s(y)) → LE(x, y)

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

POL(s(x₁)) = 1 + (4)x_1
POL(LE(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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(s(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)) = (3)x_2
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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

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

F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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.

F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)

The remaining pairs can at least be oriented weakly.

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

Used ordering: Polynomial interpretation [25,35]:

POL(F(x₁, x₂, x₃, x₄)) = x_1
POL(minus(x₁, x₂)) = 1
POL(s(x₁)) = 1 + (4)x_1
POL(0) = 2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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.

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

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

POL(F(x₁, x₂, x₃, x₄)) = (3)x_2
POL(minus(x₁, x₂)) = (2)x_1
POL(s(x₁)) = 3 + (3)x_1
POL(0) = 1

The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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.