Termination w.r.t. Q proof of /tmp/tmpu1YqCE/aprove01.srs

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:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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:

HALF(s(s(x1))) → P(s(s(x1)))
HALF(s(s(x1))) → P(p(s(s(x1))))
HALF(0(x1)) → HALF(p(s(p(s(x1)))))
P(p(s(x1))) → P(x1)
RD(0(x1)) → RD(x1)
HALF(0(x1)) → P(s(p(s(x1))))
P(0(x1)) → P(x1)
G(s(x1)) → P(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
J(s(x1)) → P(s(p(p(s(x1)))))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
J(s(x1)) → F(p(s(p(p(s(x1))))))
J(s(x1)) → P(s(x1))
F(s(x1)) → G(p(s(s(x1))))
G(s(x1)) → P(s(p(s(x1))))
J(s(x1)) → P(s(f(p(s(p(p(s(x1))))))))
G(s(x1)) → J(s(p(s(p(s(x1))))))
F(s(x1)) → P(s(s(x1)))
J(s(x1)) → P(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
HALF(0(x1)) → P(s(x1))
G(s(x1)) → P(s(x1))
J(s(x1)) → P(p(s(x1)))
F(s(x1)) → P(s(g(p(s(s(x1))))))
G(s(x1)) → P(s(s(s(j(s(p(s(p(s(x1))))))))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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:

HALF(s(s(x1))) → P(s(s(x1)))
HALF(s(s(x1))) → P(p(s(s(x1))))
HALF(0(x1)) → HALF(p(s(p(s(x1)))))
P(p(s(x1))) → P(x1)
RD(0(x1)) → RD(x1)
HALF(0(x1)) → P(s(p(s(x1))))
P(0(x1)) → P(x1)
G(s(x1)) → P(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
J(s(x1)) → P(s(p(p(s(x1)))))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))
J(s(x1)) → F(p(s(p(p(s(x1))))))
J(s(x1)) → P(s(x1))
F(s(x1)) → G(p(s(s(x1))))
G(s(x1)) → P(s(p(s(x1))))
J(s(x1)) → P(s(f(p(s(p(p(s(x1))))))))
G(s(x1)) → J(s(p(s(p(s(x1))))))
F(s(x1)) → P(s(s(x1)))
J(s(x1)) → P(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
HALF(0(x1)) → P(s(x1))
G(s(x1)) → P(s(x1))
J(s(x1)) → P(p(s(x1)))
F(s(x1)) → P(s(g(p(s(s(x1))))))
G(s(x1)) → P(s(s(s(j(s(p(s(p(s(x1))))))))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

RD(0(x1)) → RD(x1)

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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.

RD(0(x1)) → RD(x1)

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

POL(RD(x₁)) = (2)x_1
POL(0(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
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:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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:

P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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.

P(p(s(x1))) → P(x1)
P(0(x1)) → P(x1)

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

POL(P(x₁)) = (3/2)x_1
POL(p(x₁)) = 3/2 + (7/4)x_1
POL(s(x₁)) = 9/4 + (3)x_1
POL(0(x₁)) = 9/4 + (7/2)x_1

The value of delta used in the strict ordering is 27/8.
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:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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:

HALF(0(x1)) → HALF(p(s(p(s(x1)))))
HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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.

HALF(0(x1)) → HALF(p(s(p(s(x1)))))

The remaining pairs can at least be oriented weakly.

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x₁)) = (1/4)x_1
POL(p(x₁)) = x_1
POL(s(x₁)) = x_1
POL(0(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:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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.

HALF(s(s(x1))) → HALF(p(p(s(s(x1)))))

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

POL(HALF(x₁)) = (4)x_1
POL(s(x₁)) = 4 + (4)x_1
POL(p(x₁)) = 4 + (1/4)x_1
POL(0(x₁)) = 1/2

The value of delta used in the strict ordering is 55.
The following usable rules [17] were oriented:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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

            ↳ QDPOrderProof

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

J(s(x1)) → F(p(s(p(p(s(x1))))))
F(s(x1)) → G(p(s(s(x1))))
G(s(x1)) → J(s(p(s(p(s(x1))))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

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(x1)) → G(p(s(s(x1))))
G(s(x1)) → J(s(p(s(p(s(x1))))))

The remaining pairs can at least be oriented weakly.

J(s(x1)) → F(p(s(p(p(s(x1))))))

Used ordering: Polynomial interpretation [25,35]:

POL(J(x₁)) = 1/4 + (1/4)x_1
POL(s(x₁)) = 1/4 + (4)x_1
POL(p(x₁)) = (1/4)x_1
POL(G(x₁)) = 1/2 + (1/2)x_1
POL(0(x₁)) = 0
POL(F(x₁)) = (4)x_1

The value of delta used in the strict ordering is 3/16.
The following usable rules [17] were oriented:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(p(x1))))
p(p(s(x1))) → p(x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

J(s(x1)) → F(p(s(p(p(s(x1))))))

The TRS R consists of the following rules:

p(0(x1)) → 0(s(s(p(x1))))
p(s(x1)) → x1
p(p(s(x1))) → p(x1)
f(s(x1)) → p(s(g(p(s(s(x1))))))
g(s(x1)) → p(p(s(s(s(j(s(p(s(p(s(x1)))))))))))
j(s(x1)) → p(s(s(p(s(f(p(s(p(p(s(x1)))))))))))
half(0(x1)) → 0(s(s(half(p(s(p(s(x1))))))))
half(s(s(x1))) → s(half(p(p(s(s(x1))))))
rd(0(x1)) → 0(s(0(0(0(0(s(0(rd(x1)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.