Termination w.r.t. Q proof of /tmp/tmpqXa0tB/z094.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:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(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:

F(x1) → C(c(x1))
C(f(x1)) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
N(f(x1)) → F(n(x1))
C(f(x1)) → C(x1)
F(x1) → C(x1)
N(s(x1)) → F(s(s(x1)))
N(f(x1)) → N(x1)
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(x1) → C(c(x1))
C(f(x1)) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
N(f(x1)) → F(n(x1))
C(f(x1)) → C(x1)
F(x1) → C(x1)
N(s(x1)) → F(s(s(x1)))
N(f(x1)) → N(x1)
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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:

C(f(x1)) → C(c(x1))
C(f(x1)) → F(c(c(x1)))
C(f(x1)) → C(x1)
N(f(x1)) → N(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(C(x₁)) = x₁
POL(F(x₁)) = x₁
POL(N(x₁)) = x₁
POL(c(x₁)) = x₁
POL(f(x₁)) = 1 + x₁
POL(n(x₁)) = 1 + x₁
POL(s(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(x1) → C(c(x1))
N(f(x1)) → F(n(x1))
F(x1) → C(x1)
N(s(x1)) → F(s(s(x1)))
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N(f(x1)) → F(n(x1))
F(x1) → N(c(c(x1)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(x1) → N(c(c(x1))) at position [0] we obtained the following new rules:

F(c(x0)) → N(c(c(x0)))
F(f(x0)) → N(c(f(c(c(x0)))))
F(x0) → N(c(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(x0) → N(c(x0))
N(f(x1)) → F(n(x1))
F(f(x0)) → N(c(f(c(c(x0)))))
F(c(x0)) → N(c(c(x0)))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(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(c(x0)) → N(c(c(x0)))

The remaining pairs can at least be oriented weakly.

F(x0) → N(c(x0))
N(f(x1)) → F(n(x1))
F(f(x0)) → N(c(f(c(c(x0)))))

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

POL( N(x₁) ) = max{0, -1}

POL( s(x₁) ) = max{0, x₁ - 1}

POL( n(x₁) ) = max{0, -1}

POL( f(x₁) ) = max{0, -1}

POL( c(x₁) ) = 1

POL( F(x₁) ) = x₁

The following usable rules [17] were oriented:

n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(s(x1)))
f(x1) → n(c(c(x1)))
c(c(x1)) → c(x1)
c(f(x1)) → f(c(c(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ SemLabProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(x0) → N(c(x0))
N(f(x1)) → F(n(x1))
F(f(x0)) → N(c(f(c(c(x0)))))

The TRS R consists of the following rules:

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.N: 0
c: 0
n: x₀
f: x₀
s: 1
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.1(x0) → N.0(c.1(x0))
N.0(f.0(x1)) → F.0(n.0(x1))
N.1(f.1(x1)) → F.0(n.1(x1))
F.1(f.1(x0)) → N.0(c.0(f.0(c.0(c.1(x0)))))
F.0(f.0(x0)) → N.0(c.0(f.0(c.0(c.0(x0)))))
N.1(f.1(x1)) → F.1(n.1(x1))
F.0(x0) → N.0(c.0(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
c.0(c.0(x1)) → c.0(x1)
f.1(x1) → n.0(c.0(c.1(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))
f.0(x1) → n.0(c.0(c.0(x1)))
c.0(f.0(x1)) → f.0(c.0(c.0(x1)))
n.1(s.0(x1)) → f.1(s.1(s.0(x1)))
c.0(c.1(x1)) → c.1(x1)
n.1(f.1(x1)) → f.1(n.1(x1))
n.1(x0) → n.0(x0)
n.1(s.1(x1)) → f.1(s.1(s.1(x1)))
s.1(x0) → s.0(x0)
f.1(x0) → f.0(x0)
c.1(f.1(x1)) → f.0(c.0(c.1(x1)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F.1(x0) → N.0(c.1(x0))
N.0(f.0(x1)) → F.0(n.0(x1))
N.1(f.1(x1)) → F.0(n.1(x1))
F.1(f.1(x0)) → N.0(c.0(f.0(c.0(c.1(x0)))))
F.0(f.0(x0)) → N.0(c.0(f.0(c.0(c.0(x0)))))
N.1(f.1(x1)) → F.1(n.1(x1))
F.0(x0) → N.0(c.0(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
c.0(c.0(x1)) → c.0(x1)
f.1(x1) → n.0(c.0(c.1(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))
f.0(x1) → n.0(c.0(c.0(x1)))
c.0(f.0(x1)) → f.0(c.0(c.0(x1)))
n.1(s.0(x1)) → f.1(s.1(s.0(x1)))
c.0(c.1(x1)) → c.1(x1)
n.1(f.1(x1)) → f.1(n.1(x1))
n.1(x0) → n.0(x0)
n.1(s.1(x1)) → f.1(s.1(s.1(x1)))
s.1(x0) → s.0(x0)
f.1(x0) → f.0(x0)
c.1(f.1(x1)) → f.0(c.0(c.1(x1)))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N.0(f.0(x1)) → F.0(n.0(x1))
F.0(f.0(x0)) → N.0(c.0(f.0(c.0(c.0(x0)))))
F.0(x0) → N.0(c.0(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
c.0(c.0(x1)) → c.0(x1)
f.1(x1) → n.0(c.0(c.1(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))
f.0(x1) → n.0(c.0(c.0(x1)))
c.0(f.0(x1)) → f.0(c.0(c.0(x1)))
n.1(s.0(x1)) → f.1(s.1(s.0(x1)))
c.0(c.1(x1)) → c.1(x1)
n.1(f.1(x1)) → f.1(n.1(x1))
n.1(x0) → n.0(x0)
n.1(s.1(x1)) → f.1(s.1(s.1(x1)))
s.1(x0) → s.0(x0)
f.1(x0) → f.0(x0)
c.1(f.1(x1)) → f.0(c.0(c.1(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

c.1(f.1(x1)) → f.0(c.0(c.1(x1)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0(x₁)) = x₁
POL(N.0(x₁)) = x₁
POL(c.0(x₁)) = x₁
POL(c.1(x₁)) = x₁
POL(f.0(x₁)) = x₁
POL(f.1(x₁)) = 1 + x₁
POL(n.0(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N.0(f.0(x1)) → F.0(n.0(x1))
F.0(f.0(x0)) → N.0(c.0(f.0(c.0(c.0(x0)))))
F.0(x0) → N.0(c.0(x0))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
c.0(c.0(x1)) → c.0(x1)
c.0(c.1(x1)) → c.1(x1)
c.0(f.0(x1)) → f.0(c.0(c.0(x1)))
f.0(x1) → n.0(c.0(c.0(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))

c.1(x0) → c.0(x0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0(x₁)) = x₁
POL(N.0(x₁)) = x₁
POL(c.0(x₁)) = x₁
POL(c.1(x₁)) = 1 + x₁
POL(f.0(x₁)) = x₁
POL(n.0(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

N.0(f.0(x1)) → F.0(n.0(x1))
F.0(f.0(x0)) → N.0(c.0(f.0(c.0(c.0(x0)))))
F.0(x0) → N.0(c.0(x0))

The TRS R consists of the following rules:

c.0(c.0(x1)) → c.0(x1)
c.0(c.1(x1)) → c.1(x1)
c.0(f.0(x1)) → f.0(c.0(c.0(x1)))
f.0(x1) → n.0(c.0(c.0(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))

N.0(f.0(x1)) → F.0(n.0(x1))

Strictly oriented rules of the TRS R:

f.0(x1) → n.0(c.0(c.0(x1)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0(x₁)) = x₁
POL(N.0(x₁)) = x₁
POL(c.0(x₁)) = x₁
POL(c.1(x₁)) = x₁
POL(f.0(x₁)) = 1 + x₁
POL(n.0(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F.0(f.0(x0)) → N.0(c.0(f.0(c.0(c.0(x0)))))
F.0(x0) → N.0(c.0(x0))

The TRS R consists of the following rules:

c.0(c.0(x1)) → c.0(x1)
c.0(c.1(x1)) → c.1(x1)
c.0(f.0(x1)) → f.0(c.0(c.0(x1)))
n.0(f.0(x1)) → f.0(n.0(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 2 less nodes.
We have reversed the following QTRS:
The set of rules R is

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

The set Q is empty.
We have obtained the following QTRS:

f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

f(x1) → n(c(c(x1)))
c(f(x1)) → f(c(c(x1)))
c(c(x1)) → c(x1)
n(s(x1)) → f(s(s(x1)))
n(f(x1)) → f(n(x1))

The set Q is empty.
We have obtained the following QTRS:

f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))

Q is empty.