Termination w.r.t. Q proof of /tmp/tmpaF2JMg/aprove07.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:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

Q is empty.

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

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

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

0(foo(x)) → s(p(s(s(s(p(p(p(s(0(x))))))))))
s(foo(x)) → s(p(s(s(p(p(bar(s(p(s(s(p(p(foo(s(p(s(s(p(s(s(p(p(p(s(p(x))))))))))))))))))))))))))
0(bar(x)) → s(s(s(p(0(x)))))
s(bar(x)) → s(s(p(p(s(foo(s(s(p(p(s(p(x))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(0(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:

0(foo(x)) → s(p(s(s(s(p(p(p(s(0(x))))))))))
s(foo(x)) → s(p(s(s(p(p(bar(s(p(s(s(p(p(foo(s(p(s(s(p(s(s(p(p(p(s(p(x))))))))))))))))))))))))))
0(bar(x)) → s(s(s(p(0(x)))))
s(bar(x)) → s(s(p(p(s(foo(s(s(p(p(s(p(x))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(0(x)))))

Q is empty.

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

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

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

0(foo(x)) → s(p(s(s(s(p(p(p(s(0(x))))))))))
s(foo(x)) → s(p(s(s(p(p(bar(s(p(s(s(p(p(foo(s(p(s(s(p(s(s(p(p(p(s(p(x))))))))))))))))))))))))))
0(bar(x)) → s(s(s(p(0(x)))))
s(bar(x)) → s(s(p(p(s(foo(s(s(p(p(s(p(x))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(0(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:

0(foo(x)) → s(p(s(s(s(p(p(p(s(0(x))))))))))
s(foo(x)) → s(p(s(s(p(p(bar(s(p(s(s(p(p(foo(s(p(s(s(p(s(s(p(p(p(s(p(x))))))))))))))))))))))))))
0(bar(x)) → s(s(s(p(0(x)))))
s(bar(x)) → s(s(p(p(s(foo(s(s(p(p(s(p(x))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(0(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:

FOO(s(x1)) → P(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))
FOO(0(x1)) → P(p(p(s(s(s(p(s(x1))))))))
FOO(s(x1)) → P(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))
FOO(s(x1)) → P(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))
FOO(s(x1)) → P(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))
BAR(s(x1)) → P(p(s(s(x1))))
FOO(s(x1)) → P(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(x1)))))
FOO(s(x1)) → P(s(bar(p(p(s(s(p(s(x1)))))))))
P(p(s(x1))) → P(x1)
BAR(s(x1)) → P(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
FOO(0(x1)) → P(p(s(s(s(p(s(x1)))))))
FOO(0(x1)) → P(s(x1))
FOO(s(x1)) → P(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))
FOO(s(x1)) → P(p(s(s(p(s(x1))))))
BAR(0(x1)) → P(s(s(s(x1))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))
FOO(s(x1)) → P(s(x1))
FOO(s(x1)) → P(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))
FOO(0(x1)) → P(s(s(s(p(s(x1))))))
BAR(s(x1)) → P(p(s(s(foo(s(p(p(s(s(x1))))))))))
FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
BAR(s(x1)) → P(s(s(foo(s(p(p(s(s(x1)))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → P(s(s(x1)))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

FOO(s(x1)) → P(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))
FOO(0(x1)) → P(p(p(s(s(s(p(s(x1))))))))
FOO(s(x1)) → P(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))
FOO(s(x1)) → P(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))
FOO(s(x1)) → P(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))
BAR(s(x1)) → P(p(s(s(x1))))
FOO(s(x1)) → P(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(x1)))))
FOO(s(x1)) → P(s(bar(p(p(s(s(p(s(x1)))))))))
P(p(s(x1))) → P(x1)
BAR(s(x1)) → P(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
FOO(0(x1)) → P(p(s(s(s(p(s(x1)))))))
FOO(0(x1)) → P(s(x1))
FOO(s(x1)) → P(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))
FOO(s(x1)) → P(p(s(s(p(s(x1))))))
BAR(0(x1)) → P(s(s(s(x1))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))
FOO(s(x1)) → P(s(x1))
FOO(s(x1)) → P(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))
FOO(0(x1)) → P(s(s(s(p(s(x1))))))
BAR(s(x1)) → P(p(s(s(foo(s(p(p(s(s(x1))))))))))
FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
BAR(s(x1)) → P(s(s(foo(s(p(p(s(s(x1)))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → P(s(s(x1)))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

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

P(p(s(x1))) → P(x1)

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

            ↳ UsableRulesProof

          ↳ QDP

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

P(p(s(x1))) → P(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:

P(p(s(x1))) → P(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x₁)) = 2·x₁
POL(p(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

            ↳ UsableRulesProof

          ↳ 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.
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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

          ↳ QDP

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

P(p(s(x1))) → P(x1)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

            ↳ MNOCProof

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ MNOCProof

            ↳ MNOCProof

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

            ↳ MNOCProof

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

Q is empty.
We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))) at position [0,0] we obtained the following new rules:

FOO(s(x1)) → FOO(p(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

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

FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))
FOO(s(x1)) → FOO(p(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → BAR(p(p(s(s(p(s(x1))))))) at position [0,0] we obtained the following new rules:

FOO(s(x1)) → BAR(p(s(p(s(x1)))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

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

BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))
FOO(s(x1)) → FOO(p(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))
FOO(s(x1)) → BAR(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule BAR(s(x1)) → FOO(s(p(p(s(s(x1)))))) at position [0,0,0] we obtained the following new rules:

BAR(s(x1)) → FOO(s(p(s(x1))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

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

FOO(s(x1)) → FOO(p(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))
FOO(s(x1)) → BAR(p(s(p(s(x1)))))
BAR(s(x1)) → FOO(s(p(s(x1))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → FOO(p(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))) at position [0] we obtained the following new rules:

FOO(s(x1)) → FOO(p(s(bar(p(p(s(s(p(s(x1))))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

FOO(s(x1)) → FOO(p(s(bar(p(p(s(s(p(s(x1))))))))))
FOO(s(x1)) → BAR(p(s(p(s(x1)))))
BAR(s(x1)) → FOO(s(p(s(x1))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → BAR(p(s(p(s(x1))))) at position [0] we obtained the following new rules:

FOO(s(x1)) → BAR(p(s(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

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

FOO(s(x1)) → FOO(p(s(bar(p(p(s(s(p(s(x1))))))))))
FOO(s(x1)) → BAR(p(s(x1)))
BAR(s(x1)) → FOO(s(p(s(x1))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule BAR(s(x1)) → FOO(s(p(s(x1)))) at position [0,0] we obtained the following new rules:

BAR(s(x1)) → FOO(s(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

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

FOO(s(x1)) → FOO(p(s(bar(p(p(s(s(p(s(x1))))))))))
FOO(s(x1)) → BAR(p(s(x1)))
BAR(s(x1)) → FOO(s(x1))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → FOO(p(s(bar(p(p(s(s(p(s(x1)))))))))) at position [0] we obtained the following new rules:

FOO(s(x1)) → FOO(bar(p(p(s(s(p(s(x1))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

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

FOO(s(x1)) → FOO(bar(p(p(s(s(p(s(x1))))))))
FOO(s(x1)) → BAR(p(s(x1)))
BAR(s(x1)) → FOO(s(x1))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → BAR(p(s(x1))) at position [0] we obtained the following new rules:

FOO(s(x1)) → BAR(x1)

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

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

FOO(s(x1)) → BAR(x1)
FOO(s(x1)) → FOO(bar(p(p(s(s(p(s(x1))))))))
BAR(s(x1)) → FOO(s(x1))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → FOO(bar(p(p(s(s(p(s(x1)))))))) at position [0,0,0] we obtained the following new rules:

FOO(s(x1)) → FOO(bar(p(s(p(s(x1))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

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

FOO(s(x1)) → FOO(bar(p(s(p(s(x1))))))
FOO(s(x1)) → BAR(x1)
BAR(s(x1)) → FOO(s(x1))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → FOO(bar(p(s(p(s(x1)))))) at position [0,0] we obtained the following new rules:

FOO(s(x1)) → FOO(bar(p(s(x1))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

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

FOO(s(x1)) → BAR(x1)
FOO(s(x1)) → FOO(bar(p(s(x1))))
BAR(s(x1)) → FOO(s(x1))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FOO(s(x1)) → FOO(bar(p(s(x1)))) at position [0,0] we obtained the following new rules:

FOO(s(x1)) → FOO(bar(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ QDPOrderProof

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

FOO(s(x1)) → BAR(x1)
FOO(s(x1)) → FOO(bar(x1))
BAR(s(x1)) → FOO(s(x1))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(x0))

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.

FOO(s(x1)) → FOO(bar(x1))
BAR(s(x1)) → FOO(s(x1))

The remaining pairs can at least be oriented weakly.

FOO(s(x1)) → BAR(x1)

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

POL( foo(x₁) ) = max{0, -1}

POL( bar(x₁) ) = max{0, -1}

POL( s(x₁) ) = x₁ + 1

POL( p(x₁) ) = max{0, x₁ - 1}

POL( 0(x₁) ) = max{0, -1}

POL( FOO(x₁) ) = x₁

POL( BAR(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

p(0(x1)) → 0(s(s(s(s(x1)))))
p(s(x1)) → x1
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ Rewriting

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Rewriting

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ QDPOrderProof

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

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

FOO(s(x1)) → BAR(x1)

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))

The set Q consists of the following terms:

foo(0(x0))
foo(s(x0))
bar(0(x0))
bar(s(x0))
p(s(x0))
p(0(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.