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:

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

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

o(f(x1)) → f(o(x1))
Used ordering:
Polynomial interpretation [25]:

POL(c(x1)) = x1   
POL(f(x1)) = 2 + x1   
POL(n(x1)) = 2 + x1   
POL(o(x1)) = 1 + 2·x1   
POL(s(x1)) = 1 + x1   
POL(t(x1)) = x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

o(s(x1)) → f(s(x1))
Used ordering:
Polynomial interpretation [25]:

POL(c(x1)) = x1   
POL(f(x1)) = 2·x1   
POL(n(x1)) = 2·x1   
POL(o(x1)) = 1 + 2·x1   
POL(s(x1)) = 2·x1   
POL(t(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

C(n(x1)) → C(x1)
T(f(x1)) → C(n(x1))
T(f(x1)) → T(c(n(x1)))
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)
C(n(x1)) → N(c(x1))
N(f(x1)) → N(x1)
T(f(x1)) → N(x1)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
T(f(x1)) → C(n(x1))
T(f(x1)) → T(c(n(x1)))
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)
C(n(x1)) → N(c(x1))
N(f(x1)) → N(x1)
T(f(x1)) → N(x1)

The TRS R consists of the following rules:

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

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 3 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                  ↳ QDP
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

N(f(x1)) → N(x1)

The TRS R consists of the following rules:

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

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

N(f(x1)) → N(x1)

R is empty.
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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

N(f(x1)) → N(x1)

R is empty.
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(f(x1)) → N(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(N(x1)) = 2·x1   
POL(f(x1)) = 2·x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
                          ↳ QDP
                            ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
P is empty.
R is empty.
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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(o(x1)) → C(x1)
C(f(x1)) → C(x1)

The TRS R consists of the following rules:

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

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(f(x1)) → C(x1)
C(o(x1)) → C(x1)

R is empty.
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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(n(x1)) → C(x1)
C(f(x1)) → C(x1)
C(o(x1)) → C(x1)

R is empty.
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:

C(n(x1)) → C(x1)
C(f(x1)) → C(x1)
C(o(x1)) → C(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(C(x1)) = 2·x1   
POL(f(x1)) = 2·x1   
POL(n(x1)) = 2·x1   
POL(o(x1)) = 2·x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
                          ↳ QDP
                            ↳ PisEmptyProof
                  ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

Q DP problem:
P is empty.
R is empty.
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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

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

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                    ↳ UsableRulesProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

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

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
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

T(f(x1)) → T(c(n(x1)))

The TRS R consists of the following rules:

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

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

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))
T(f(s(x0))) → T(c(f(s(x0))))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))
T(f(s(x0))) → T(c(f(s(x0))))

The TRS R consists of the following rules:

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


T(f(s(x0))) → T(c(f(s(x0))))
The remaining pairs can at least be oriented weakly.

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( s(x1) ) = x1 + 1


POL( T(x1) ) = x1


POL( f(x1) ) = x1


POL( n(x1) ) = x1


POL( c(x1) ) = max{0, -1}


POL( o(x1) ) = max{0, -1}



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ SemLabProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

T(f(f(x0))) → T(c(f(n(x0))))
T(f(x0)) → T(n(c(x0)))

The TRS R consists of the following rules:

n(f(x1)) → f(n(x1))
n(s(x1)) → f(s(x1))
c(f(x1)) → f(c(x1))
c(n(x1)) → n(c(x1))
c(o(x1)) → o(c(x1))
c(o(x1)) → o(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: 1
f: x0
n: x0
T: 0
s: 0
o: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ SemLabProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

c.0(n.0(x1)) → n.1(c.0(x1))
n.0(s.1(x1)) → f.0(s.1(x1))
n.1(f.1(x1)) → f.1(n.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(f.1(x1)) → f.1(c.1(x1))
n.0(f.0(x1)) → f.0(n.0(x1))
c.1(o.0(x1)) → o.1(c.0(x1))
n.0(s.0(x1)) → f.0(s.0(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.0(f.0(x1)) → f.1(c.0(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(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 2 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ 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:

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

The TRS R consists of the following rules:

c.0(n.0(x1)) → n.1(c.0(x1))
n.0(s.1(x1)) → f.0(s.1(x1))
n.1(f.1(x1)) → f.1(n.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(f.1(x1)) → f.1(c.1(x1))
n.0(f.0(x1)) → f.0(n.0(x1))
c.1(o.0(x1)) → o.1(c.0(x1))
n.0(s.0(x1)) → f.0(s.0(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.0(f.0(x1)) → f.1(c.0(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(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(o.0(x1)) → o.1(c.0(x1))
c.0(n.0(x1)) → n.1(c.0(x1))
c.0(f.0(x1)) → f.1(c.0(x1))
Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x1)) = x1   
POL(c.0(x1)) = x1   
POL(c.1(x1)) = x1   
POL(f.0(x1)) = 1 + x1   
POL(f.1(x1)) = x1   
POL(n.0(x1)) = 1 + x1   
POL(n.1(x1)) = x1   
POL(o.0(x1)) = 1 + x1   
POL(o.1(x1)) = x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ 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:

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

The TRS R consists of the following rules:

n.1(f.1(x1)) → f.1(n.1(x1))
c.1(f.1(x1)) → f.1(c.1(x1))
c.1(o.1(x1)) → o.1(x1)
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(o.0(x1)) → o.0(x1)
c.1(o.1(x1)) → o.1(c.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 rules of the TRS R:

c.1(o.1(x1)) → o.1(x1)
c.1(o.0(x1)) → o.0(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x1)) = x1   
POL(c.1(x1)) = 1 + x1   
POL(f.1(x1)) = 1 + x1   
POL(n.1(x1)) = x1   
POL(o.0(x1)) = x1   
POL(o.1(x1)) = x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ 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:

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))

The TRS R consists of the following rules:

n.1(f.1(x1)) → f.1(n.1(x1))
c.1(f.1(x1)) → f.1(c.1(x1))
c.1(n.1(x1)) → n.1(c.1(x1))
c.1(o.1(x1)) → o.1(c.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:

T.1(f.1(x0)) → T.1(n.1(c.1(x0)))
T.1(f.1(f.1(x0))) → T.1(c.1(f.1(n.1(x0))))


Used ordering: POLO with Polynomial interpretation [25]:

POL(T.1(x1)) = x1   
POL(c.1(x1)) = x1   
POL(f.1(x1)) = 1 + x1   
POL(n.1(x1)) = x1   
POL(o.1(x1)) = x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ SemLabProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ UsableRulesReductionPairsProof
                                          ↳ QDP
                                            ↳ RuleRemovalProof
                                              ↳ QDP
                                                ↳ RuleRemovalProof
                                                  ↳ QDP
                                                    ↳ PisEmptyProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.