Termination w.r.t. Q proof of /tmp/tmptD2Kly/Liveness

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:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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:

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → F(X)
TP(mark(x)) → F(f(f(X)))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) → F(f(f(X)))
CHK(no(c)) → ACTIVE(c)
CHK(no(f(x))) → F(f(f(f(X))))
TP(mark(x)) → F(f(X))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
F(no(x)) → F(x)
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(X))))))))
F(active(x)) → F(x)
F(mark(x)) → F(x)
MAT(f(x), f(y)) → F(mat(x, y))
CHK(no(f(x))) → F(f(f(f(f(X)))))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) → F(f(X))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))
TP(mark(x)) → F(f(f(f(X))))
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
TP(mark(x)) → F(f(f(f(f(f(X))))))
TP(mark(x)) → F(f(f(f(f(X)))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → F(f(f(f(f(f(X))))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(X)
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(X))))))))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

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:

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) → F(X)
TP(mark(x)) → F(f(f(X)))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) → F(f(f(X)))
CHK(no(c)) → ACTIVE(c)
CHK(no(f(x))) → F(f(f(f(X))))
TP(mark(x)) → F(f(X))
CHK(no(f(x))) → F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
F(no(x)) → F(x)
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) → F(f(f(f(f(f(f(f(X))))))))
F(active(x)) → F(x)
F(mark(x)) → F(x)
MAT(f(x), f(y)) → F(mat(x, y))
CHK(no(f(x))) → F(f(f(f(f(X)))))
TP(mark(x)) → F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) → F(f(X))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))
TP(mark(x)) → F(f(f(f(X))))
CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
TP(mark(x)) → F(f(f(f(f(f(X))))))
TP(mark(x)) → F(f(f(f(f(X)))))
CHK(no(f(x))) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) → F(f(f(f(f(f(X))))))
TP(mark(x)) → MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) → F(X)
MAT(f(x), f(y)) → MAT(x, y)
TP(mark(x)) → F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) → F(f(f(f(f(f(f(f(X))))))))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(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 3 SCCs with 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)
F(no(x)) → F(x)
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))
F(active(x)) → F(x)

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)
ACTIVE(f(x)) → F(f(x))
F(no(x)) → F(x)
F(active(x)) → ACTIVE(f(x))
F(active(x)) → F(x)

The TRS R consists of the following rules:

f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(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:

F(active(x)) → F(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = 1 + x₁
POL(F(x₁)) = 1 + x₁
POL(active(x₁)) = 1 + 2·x₁
POL(f(x₁)) = 1 + 2·x₁
POL(mark(x₁)) = x₁
POL(no(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)
F(no(x)) → F(x)
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))

The TRS R consists of the following rules:

f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))

F(no(x)) → F(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(F(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(f(x₁)) = x₁
POL(mark(x₁)) = 2·x₁
POL(no(x₁)) = 1 + 2·x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)
ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))

The TRS R consists of the following rules:

f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))

ACTIVE(f(x)) → F(f(x))
F(active(x)) → ACTIVE(f(x))

Strictly oriented rules of the TRS R:

f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
active(f(x)) → mark(f(f(x)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = 1 + 2·x₁
POL(F(x₁)) = 2·x₁
POL(active(x₁)) = 2 + 2·x₁
POL(f(x₁)) = 1 + 2·x₁
POL(mark(x₁)) = x₁
POL(no(x₁)) = 1 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ MNOCProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)

The TRS R consists of the following rules:

f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ MNOCProof

                              ↳ QDP

                                ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)

The TRS R consists of the following rules:

f(mark(x)) → mark(f(x))

The set Q consists of the following terms:

f(mark(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ MNOCProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)

R is empty.
The set Q consists of the following terms:

f(mark(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(mark(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ MNOCProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

F(mark(x)) → F(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

F(mark(x)) → F(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

          ↳ QDP

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

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ RuleRemovalProof

          ↳ QDP

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

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)

The set Q consists of the following terms:

mat(f(x0), f(y))
mat(f(x0), c)

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:

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

Strictly oriented rules of the TRS R:

mat(f(x), c) → no(c)

Used ordering: POLO with Polynomial interpretation [25]:

POL(CHK(x₁)) = x₁
POL(X) = 0
POL(c) = 2
POL(f(x₁)) = 2·x₁
POL(mat(x₁, x₂)) = x₁ + 2·x₂
POL(no(x₁)) = 1 + x₁
POL(y) = 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

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

mat(f(x), f(y)) → f(mat(x, y))

The set Q consists of the following terms:

mat(f(x0), f(y))
mat(f(x0), c)

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

            ↳ UsableRulesProof

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

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following rules:

active(f(x)) → mark(f(f(x)))
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
mat(f(x), c) → no(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
tp(mark(x)) → tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ Narrowing

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

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) → active(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))) at position [0] we obtained the following new rules:

TP(mark(f(y))) → TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
TP(mark(c)) → TP(chk(no(c)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

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

TP(mark(c)) → TP(chk(no(c)))
TP(mark(f(y))) → TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) → active(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(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.