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(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

Q is empty.

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

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

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

0(f(x)) → 0(s(x))
0(d(x)) → 0(x)
s(d(x)) → d(s(s(x)))
s(f(x)) → f(d(x))

The set Q is empty.

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

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

0(f(x)) → 0(s(x))
0(d(x)) → 0(x)
s(d(x)) → d(s(s(x)))
s(f(x)) → f(d(x))

Q is empty.

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

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

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

0(f(x)) → 0(s(x))
0(d(x)) → 0(x)
s(d(x)) → d(s(s(x)))
s(f(x)) → f(d(x))

The set Q is empty.

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

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

0(f(x)) → 0(s(x))
0(d(x)) → 0(x)
s(d(x)) → d(s(s(x)))
s(f(x)) → f(d(x))

Q is empty.

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

f(0(x1)) → s(0(x1))
d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(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:

f(0(x1)) → s(0(x1))
Used ordering:
Polynomial interpretation [25]:

POL(0(x1)) = 1 + 2·x1   
POL(d(x1)) = x1   
POL(f(x1)) = 2·x1   
POL(s(x1)) = x1   




↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence

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

d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof

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

d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

The set Q consists of the following terms:

d(0(x0))
d(s(x0))
f(s(x0))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

D(s(x1)) → D(x1)
F(s(x1)) → F(x1)
F(s(x1)) → D(f(x1))

The TRS R consists of the following rules:

d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

The set Q consists of the following terms:

d(0(x0))
d(s(x0))
f(s(x0))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof

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

D(s(x1)) → D(x1)
F(s(x1)) → F(x1)
F(s(x1)) → D(f(x1))

The TRS R consists of the following rules:

d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

The set Q consists of the following terms:

d(0(x0))
d(s(x0))
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                  ↳ QDP

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

D(s(x1)) → D(x1)

The TRS R consists of the following rules:

d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

The set Q consists of the following terms:

d(0(x0))
d(s(x0))
f(s(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                    ↳ UsableRulesProof
                  ↳ QDP

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

D(s(x1)) → D(x1)

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

d(0(x0))
d(s(x0))
f(s(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.

d(0(x0))
d(s(x0))
f(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP

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

D(s(x1)) → D(x1)

R is empty.
Q is empty.
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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                  ↳ QDP

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

D(s(x1)) → D(x1)

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

d(0(x0))
d(s(x0))
f(s(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.

d(0(x0))
d(s(x0))
f(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
                  ↳ QDP

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

D(s(x1)) → D(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:

D(s(x1)) → D(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(D(x1)) = 2·x1   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
                              ↳ QDP
                                ↳ PisEmptyProof
                  ↳ QDP

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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof

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

F(s(x1)) → F(x1)

The TRS R consists of the following rules:

d(0(x1)) → 0(x1)
d(s(x1)) → s(s(d(x1)))
f(s(x1)) → d(f(x1))

The set Q consists of the following terms:

d(0(x0))
d(s(x0))
f(s(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                    ↳ UsableRulesProof

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

F(s(x1)) → F(x1)

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

d(0(x0))
d(s(x0))
f(s(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.

d(0(x0))
d(s(x0))
f(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
                    ↳ UsableRulesProof

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

F(s(x1)) → F(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:

F(s(x1)) → F(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = 2·x1   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
                              ↳ QDP
                                ↳ PisEmptyProof
                    ↳ UsableRulesProof

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.
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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof

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

F(s(x1)) → F(x1)

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

d(0(x0))
d(s(x0))
f(s(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.

d(0(x0))
d(s(x0))
f(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QReductionProof
                          ↳ QDP

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

F(s(x1)) → F(x1)

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