Termination w.r.t. Q proof of /tmp/tmpiXNBTQ/Ex25_Luc06

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:

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.

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

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

h(X) → c(n__d(X))

Used ordering:
Polynomial interpretation [25]:

POL(activate(x₁)) = x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(f(x₁)) = 2·x₁
POL(g(x₁)) = x₁
POL(h(x₁)) = 2 + x₁
POL(n__d(x₁)) = x₁
POL(n__f(x₁)) = 2·x₁
POL(n__g(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

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:

ACTIVATE(n__d(X)) → D(X)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__g(X)) → G(X)
F(f(X)) → C(n__f(n__g(n__f(X))))
C(X) → ACTIVATE(X)
C(X) → D(activate(X))

The TRS R consists of the following rules:

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

ACTIVATE(n__d(X)) → D(X)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__g(X)) → G(X)
F(f(X)) → C(n__f(n__g(n__f(X))))
C(X) → ACTIVATE(X)
C(X) → D(activate(X))

The TRS R consists of the following rules:

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

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

ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)
F(f(X)) → C(n__f(n__g(n__f(X))))
C(X) → ACTIVATE(X)

The TRS R consists of the following rules:

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

ACTIVATE(n__f(X)) → ACTIVATE(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVATE(x₁)) = x₁
POL(C(x₁)) = x₁
POL(F(x₁)) = 2 + 2·x₁
POL(activate(x₁)) = x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(f(x₁)) = 2 + 2·x₁
POL(g(x₁)) = x₁
POL(n__d(x₁)) = x₁
POL(n__f(x₁)) = 2 + 2·x₁
POL(n__g(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ QDPOrderProof

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

ACTIVATE(n__f(X)) → F(activate(X))
F(f(X)) → C(n__f(n__g(n__f(X))))
C(X) → ACTIVATE(X)

The TRS R consists of the following rules:

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

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.

ACTIVATE(n__f(X)) → F(activate(X))

The remaining pairs can at least be oriented weakly.

F(f(X)) → C(n__f(n__g(n__f(X))))
C(X) → ACTIVATE(X)

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( C(x₁) ) = x₁

POL( f(x₁) ) = x₁ + 1

POL( c(x₁) ) = max{0, -1}

POL( n__g(x₁) ) = max{0, -1}

POL( g(x₁) ) = 0

POL( n__f(x₁) ) = x₁ + 1

POL( activate(x₁) ) = x₁

POL( d(x₁) ) = 0

POL( n__d(x₁) ) = 0

POL( ACTIVATE(x₁) ) = x₁

POL( F(x₁) ) = x₁

The following usable rules [17] were oriented:

activate(X) → X
g(X) → n__g(X)
f(X) → n__f(X)
c(X) → d(activate(X))
f(f(X)) → c(n__f(n__g(n__f(X))))
activate(n__d(X)) → d(X)
activate(n__g(X)) → g(X)
activate(n__f(X)) → f(activate(X))
d(X) → n__d(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

F(f(X)) → C(n__f(n__g(n__f(X))))
C(X) → ACTIVATE(X)

The TRS R consists of the following rules:

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

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 2 less nodes.