Termination w.r.t. Q proof of /tmp/tmpXfz

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

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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 reversed the following QTRS:
The set of rules R is

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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) → a(n(c(n(x))))
f(c(x)) → c(a(n(f(x))))
a(n(x)) → c(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

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

f(x) → a(n(c(n(x))))
f(c(x)) → c(a(n(f(x))))
a(n(x)) → c(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(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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) → a(n(c(n(x))))
f(c(x)) → c(a(n(f(x))))
a(n(x)) → c(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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.

N(s(x1)) → F(s(s(x1)))

The remaining pairs can at least be oriented weakly.

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

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

POL( C(x₁) ) = 0

POL( N(x₁) ) = x₁

POL( f(x₁) ) = x₁

POL( n(x₁) ) = x₁

POL( c(x₁) ) = max{0, -1}

POL( a(x₁) ) = max{0, -1}

POL( s(x₁) ) = 1

POL( F(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

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

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

The TRS R consists of the following rules:

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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) → C(n(a(x1))) at position [0] we obtained the following new rules:

F(x0) → C(c(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

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

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

The TRS R consists of the following rules:

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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(n(a(x1)))) at position [0] we obtained the following new rules:

F(x0) → N(c(c(x0)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                  ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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 model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.C: 0
N: 0
c: 0
n: 0
f: 0
a: 1
s: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                  ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

c.0(c.0(x1)) → c.0(x1)
f.0(x1) → n.0(c.0(n.1(a.0(x1))))
f.1(x1) → n.0(c.0(n.1(a.1(x1))))
n.0(f.0(x1)) → f.0(n.0(x1))
n.0(s.1(x1)) → f.0(s.0(s.1(x1)))
c.0(f.0(x1)) → f.0(n.1(a.0(c.0(x1))))
c.0(f.1(x1)) → f.0(n.1(a.0(c.1(x1))))
n.0(s.0(x1)) → f.0(s.0(s.0(x1)))
n.1(a.0(x1)) → c.0(x1)
n.0(f.1(x1)) → f.0(n.1(x1))
c.0(c.1(x1)) → c.1(x1)
n.1(a.1(x1)) → 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 5 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                  ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

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

The following dependency pairs can be deleted:

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

The following rules are removed from R:

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(C.0(x₁)) = x₁
POL(F.0(x₁)) = x₁
POL(N.0(x₁)) = x₁
POL(N.1(x₁)) = x₁
POL(a.0(x₁)) = x₁
POL(a.1(x₁)) = 1 + 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
POL(n.1(x₁)) = x₁
POL(s.0(x₁)) = x₁
POL(s.1(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ RuleRemovalProof

                  ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

c.0(c.0(x1)) → c.0(x1)
c.0(f.0(x1)) → f.0(n.1(a.0(c.0(x1))))
c.0(c.1(x1)) → c.1(x1)
n.1(a.0(x1)) → c.0(x1)
f.0(x1) → n.0(c.0(n.1(a.0(x1))))
n.0(s.0(x1)) → f.0(s.0(s.0(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))
n.0(s.1(x1)) → f.0(s.0(s.1(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:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ RuleRemovalProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                  ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

c.0(c.0(x1)) → c.0(x1)
c.0(f.0(x1)) → f.0(n.1(a.0(c.0(x1))))
c.0(c.1(x1)) → c.1(x1)
n.1(a.0(x1)) → c.0(x1)
f.0(x1) → n.0(c.0(n.1(a.0(x1))))
n.0(s.0(x1)) → f.0(s.0(s.0(x1)))
n.0(f.0(x1)) → f.0(n.0(x1))
n.0(s.1(x1)) → f.0(s.0(s.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 3 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ RuleRemovalProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                  ↳ SemLabProof2

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                  ↳ SemLabProof2

                    ↳ QDP

                      ↳ SemLabProof

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

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

The TRS R consists of the following rules:

f(x1) → n(c(n(a(x1))))
c(f(x1)) → f(n(a(c(x1))))
n(a(x1)) → 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
a: 0
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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                  ↳ SemLabProof2

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

f.0(x1) → n.0(c.0(n.0(a.0(x1))))
c.1(x0) → c.0(x0)
c.0(c.0(x1)) → c.0(x1)
n.0(f.0(x1)) → f.0(n.0(x1))
n.0(a.1(x1)) → c.1(x1)
c.1(f.1(x1)) → f.0(n.0(a.0(c.1(x1))))
a.1(x0) → a.0(x0)
n.1(s.0(x1)) → f.1(s.1(s.0(x1)))
c.0(c.1(x1)) → c.1(x1)
n.1(x0) → n.0(x0)
n.1(f.1(x1)) → f.1(n.1(x1))
n.0(a.0(x1)) → c.0(x1)
n.1(s.1(x1)) → f.1(s.1(s.1(x1)))
f.1(x1) → n.0(c.0(n.0(a.1(x1))))
s.1(x0) → s.0(x0)
f.1(x0) → f.0(x0)
c.0(f.0(x1)) → f.0(n.0(a.0(c.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 1 SCC with 3 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                  ↳ SemLabProof2

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

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

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

The TRS R consists of the following rules:

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

F.0(x0) → N.0(c.0(c.0(x0)))

The following rules are removed from R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ SemLabProof

                  ↳ SemLabProof2

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

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

N.0(f.0(x1)) → F.0(n.0(x1))

The TRS R consists of the following rules:

c.1(x0) → c.0(x0)
c.0(c.0(x1)) → c.0(x1)
n.0(a.0(x1)) → c.0(x1)
n.0(f.0(x1)) → f.0(n.0(x1))
c.0(f.0(x1)) → f.0(n.0(a.0(c.0(x1))))
c.0(c.1(x1)) → 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 0 SCCs with 1 less node.