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(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

Q is empty.


QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse

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

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

Q is empty.

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

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

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) → 0
Used ordering:
Polynomial interpretation [25]:

POL(0) = 1   
POL(f(x1)) = 1 + x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   




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

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

f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x

Q is empty.

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

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

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

f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
p(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:

F(s(x)) → F(p(s(x)))
F(s(x)) → P(s(x))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
p(s(x0))

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

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

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

F(s(x)) → F(p(s(x)))
F(s(x)) → P(s(x))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
p(s(x0))

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

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

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

f(s(x)) → s(s(f(p(s(x)))))
p(s(x)) → x

The set Q consists of the following terms:

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

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
p(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.

f(s(x0))



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

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

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:

p(s(x)) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   



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

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

F(s(x)) → F(p(s(x)))

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

p(s(x0))

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

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

f(s(x0))
p(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.

f(s(x0))



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

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

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

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

F(s(x)) → F(p(s(x)))

The TRS R consists of the following rules:

p(s(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

f(s(x)) → s(s(f(p(s(x)))))
f(0) → 0
p(s(x)) → x

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

s(f(x)) → s(p(f(s(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x

The set Q is empty.

↳ QTRS
  ↳ RRRPoloQTRSProof
  ↳ QTRS Reverse
QTRS

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

s(f(x)) → s(p(f(s(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x

Q is empty.