Termination w.r.t. Q proof of /tmp/tmprHJ7fw/identity.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:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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:

HALF(x) → G(h, x)
F(s(x), y) → HALF(s(x))
ID(x) → F(x, s(0))
DOUBLE(x) → G(d, x)
F(s(x), y) → DOUBLE(y)
F(s(x), y) → F(half(s(x)), double(y))
G(h, s(s(x))) → G(h, x)
G(d, s(x)) → G(d, x)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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:

HALF(x) → G(h, x)
F(s(x), y) → HALF(s(x))
ID(x) → F(x, s(0))
DOUBLE(x) → G(d, x)
F(s(x), y) → DOUBLE(y)
F(s(x), y) → F(half(s(x)), double(y))
G(h, s(s(x))) → G(h, x)
G(d, s(x)) → G(d, x)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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 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:

G(h, s(s(x))) → G(h, x)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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.

G(h, s(s(x))) → G(h, x)

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

POL(s(x₁)) = 1 + x_1
POL(G(x₁, x₂)) = (2)x_2
POL(h) = 0

The value of delta used in the strict ordering is 4.
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:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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

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

G(d, s(x)) → G(d, x)

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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.

G(d, s(x)) → G(d, x)

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

POL(d) = 0
POL(s(x₁)) = 1 + (4)x_1
POL(G(x₁, x₂)) = (4)x_2

The value of delta used in the strict ordering is 4.
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:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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

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

F(s(x), y) → F(half(s(x)), double(y))

The TRS R consists of the following rules:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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.

F(s(x), y) → F(half(s(x)), double(y))

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

POL(half(x₁)) = (1/2)x_1
POL(g(x₁, x₂)) = (1/4)x_2
POL(s(x₁)) = 1/4 + (3)x_1
POL(d) = 0
POL(h) = 0
POL(F(x₁, x₂)) = (4)x_1
POL(double(x₁)) = 4 + (13/4)x_1
POL(0) = 0

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

g(x, 0) → 0
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
half(x) → g(h, x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(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.