Termination w.r.t. Q proof of /tmp/tmpjMdBuj/z091.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:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

Q is empty.

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

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

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

0(r0(x)) → r0(0(x))
1(r0(x)) → r0(1(x))
m(r0(x)) → r0(m(x))
0(r1(x)) → r1(0(x))
1(r1(x)) → r1(1(x))
m(r1(x)) → r1(m(x))
b(r0(x)) → b(0(qr(x)))
b(r1(x)) → b(1(qr(x)))
qr(0(x)) → 0(qr(x))
qr(1(x)) → 1(qr(x))
qr(m(x)) → m(ql(x))
ql(0(x)) → 0(ql(x))
ql(1(x)) → 1(ql(x))
0(ql(b(x))) → r0(b(0(x)))
1(ql(b(x))) → r1(b(1(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(r0(x)) → r0(0(x))
1(r0(x)) → r0(1(x))
m(r0(x)) → r0(m(x))
0(r1(x)) → r1(0(x))
1(r1(x)) → r1(1(x))
m(r1(x)) → r1(m(x))
b(r0(x)) → b(0(qr(x)))
b(r1(x)) → b(1(qr(x)))
qr(0(x)) → 0(qr(x))
qr(1(x)) → 1(qr(x))
qr(m(x)) → m(ql(x))
ql(0(x)) → 0(ql(x))
ql(1(x)) → 1(ql(x))
0(ql(b(x))) → r0(b(0(x)))
1(ql(b(x))) → r1(b(1(x)))

Q is empty.

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

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

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

0(r0(x)) → r0(0(x))
1(r0(x)) → r0(1(x))
m(r0(x)) → r0(m(x))
0(r1(x)) → r1(0(x))
1(r1(x)) → r1(1(x))
m(r1(x)) → r1(m(x))
b(r0(x)) → b(0(qr(x)))
b(r1(x)) → b(1(qr(x)))
qr(0(x)) → 0(qr(x))
qr(1(x)) → 1(qr(x))
qr(m(x)) → m(ql(x))
ql(0(x)) → 0(ql(x))
ql(1(x)) → 1(ql(x))
0(ql(b(x))) → r0(b(0(x)))
1(ql(b(x))) → r1(b(1(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(r0(x)) → r0(0(x))
1(r0(x)) → r0(1(x))
m(r0(x)) → r0(m(x))
0(r1(x)) → r1(0(x))
1(r1(x)) → r1(1(x))
m(r1(x)) → r1(m(x))
b(r0(x)) → b(0(qr(x)))
b(r1(x)) → b(1(qr(x)))
qr(0(x)) → 0(qr(x))
qr(1(x)) → 1(qr(x))
qr(m(x)) → m(ql(x))
ql(0(x)) → 0(ql(x))
ql(1(x)) → 1(ql(x))
0(ql(b(x))) → r0(b(0(x)))
1(ql(b(x))) → r1(b(1(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:

B(ql(1(x1))) → R1(x1)
B(ql(1(x1))) → B(r1(x1))
R0(1(x1)) → 1¹(r0(x1))
R0(m(x1)) → R0(x1)
0¹(ql(x1)) → 0¹(x1)
R1(m(x1)) → M(r1(x1))
B(ql(0(x1))) → R0(x1)
M(qr(x1)) → M(x1)
B(ql(1(x1))) → 1¹(b(r1(x1)))
R0(b(x1)) → 0¹(b(x1))
R0(m(x1)) → M(r0(x1))
R0(1(x1)) → R0(x1)
B(ql(0(x1))) → 0¹(b(r0(x1)))
B(ql(0(x1))) → B(r0(x1))
R1(b(x1)) → 1¹(b(x1))
R1(1(x1)) → R1(x1)
R0(0(x1)) → R0(x1)
R1(m(x1)) → R1(x1)
R1(0(x1)) → 0¹(r1(x1))
1¹(qr(x1)) → 1¹(x1)
R1(0(x1)) → R1(x1)
R0(0(x1)) → 0¹(r0(x1))
1¹(ql(x1)) → 1¹(x1)
R1(1(x1)) → 1¹(r1(x1))
0¹(qr(x1)) → 0¹(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(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:

B(ql(1(x1))) → R1(x1)
B(ql(1(x1))) → B(r1(x1))
R0(1(x1)) → 1¹(r0(x1))
R0(m(x1)) → R0(x1)
0¹(ql(x1)) → 0¹(x1)
R1(m(x1)) → M(r1(x1))
B(ql(0(x1))) → R0(x1)
M(qr(x1)) → M(x1)
B(ql(1(x1))) → 1¹(b(r1(x1)))
R0(b(x1)) → 0¹(b(x1))
R0(m(x1)) → M(r0(x1))
R0(1(x1)) → R0(x1)
B(ql(0(x1))) → 0¹(b(r0(x1)))
B(ql(0(x1))) → B(r0(x1))
R1(b(x1)) → 1¹(b(x1))
R1(1(x1)) → R1(x1)
R0(0(x1)) → R0(x1)
R1(m(x1)) → R1(x1)
R1(0(x1)) → 0¹(r1(x1))
1¹(qr(x1)) → 1¹(x1)
R1(0(x1)) → R1(x1)
R0(0(x1)) → 0¹(r0(x1))
1¹(ql(x1)) → 1¹(x1)
R1(1(x1)) → 1¹(r1(x1))
0¹(qr(x1)) → 0¹(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 12 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

M(qr(x1)) → M(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

M(qr(x1)) → M(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:

M(qr(x1)) → M(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(M(x₁)) = 2·x₁
POL(qr(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

M(qr(x1)) → M(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

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

1¹(qr(x1)) → 1¹(x1)
1¹(ql(x1)) → 1¹(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

1¹(qr(x1)) → 1¹(x1)
1¹(ql(x1)) → 1¹(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

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

1¹(qr(x1)) → 1¹(x1)
1¹(ql(x1)) → 1¹(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(1¹(x₁)) = 2·x₁
POL(ql(x₁)) = 2·x₁
POL(qr(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

0¹(ql(x1)) → 0¹(x1)
0¹(qr(x1)) → 0¹(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

0¹(ql(x1)) → 0¹(x1)
0¹(qr(x1)) → 0¹(x1)

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

0¹(ql(x1)) → 0¹(x1)
0¹(qr(x1)) → 0¹(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0¹(x₁)) = 2·x₁
POL(ql(x₁)) = 2·x₁
POL(qr(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

R1(0(x1)) → R1(x1)
R1(1(x1)) → R1(x1)
R1(m(x1)) → R1(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

R1(0(x1)) → R1(x1)
R1(1(x1)) → R1(x1)
R1(m(x1)) → R1(x1)

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

          ↳ QDP

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

R1(0(x1)) → R1(x1)
R1(1(x1)) → R1(x1)
R1(m(x1)) → R1(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 2·x₁
POL(1(x₁)) = 2·x₁
POL(R1(x₁)) = 2·x₁
POL(m(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

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

R0(1(x1)) → R0(x1)
R0(0(x1)) → R0(x1)
R0(m(x1)) → R0(x1)

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

R0(1(x1)) → R0(x1)
R0(0(x1)) → R0(x1)
R0(m(x1)) → R0(x1)

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

          ↳ QDP

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

R0(1(x1)) → R0(x1)
R0(0(x1)) → R0(x1)
R0(m(x1)) → R0(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 2·x₁
POL(1(x₁)) = 2·x₁
POL(R0(x₁)) = 2·x₁
POL(m(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

B(ql(1(x1))) → B(r1(x1))
B(ql(0(x1))) → B(r0(x1))

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(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.

B(ql(1(x1))) → B(r1(x1))

The remaining pairs can at least be oriented weakly.

B(ql(0(x1))) → B(r0(x1))

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

POL( r0(x₁) ) = x₁

POL( 1(x₁) ) = x₁ + 1

POL( m(x₁) ) = 1

POL( B(x₁) ) = x₁

POL( ql(x₁) ) = x₁

POL( qr(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁

POL( r1(x₁) ) = x₁

POL( 0(x₁) ) = x₁

The following usable rules [17] were oriented:

b(ql(1(x1))) → 1(b(r1(x1)))
0(ql(x1)) → ql(0(x1))
m(qr(x1)) → ql(m(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
1(ql(x1)) → ql(1(x1))
r1(b(x1)) → qr(1(b(x1)))
r0(b(x1)) → qr(0(b(x1)))
1(qr(x1)) → qr(1(x1))
0(qr(x1)) → qr(0(x1))
r1(0(x1)) → 0(r1(x1))
r0(m(x1)) → m(r0(x1))
r1(m(x1)) → m(r1(x1))
r1(1(x1)) → 1(r1(x1))
r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

B(ql(0(x1))) → B(r0(x1))

The TRS R consists of the following rules:

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(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.

B(ql(0(x1))) → B(r0(x1))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( r0(x₁) ) = x₁

POL( 1(x₁) ) = max{0, -1}

POL( m(x₁) ) = 0

POL( B(x₁) ) = x₁

POL( ql(x₁) ) = x₁

POL( qr(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁

POL( r1(x₁) ) = x₁

POL( 0(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

b(ql(1(x1))) → 1(b(r1(x1)))
0(ql(x1)) → ql(0(x1))
m(qr(x1)) → ql(m(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
1(ql(x1)) → ql(1(x1))
r1(b(x1)) → qr(1(b(x1)))
r0(b(x1)) → qr(0(b(x1)))
1(qr(x1)) → qr(1(x1))
0(qr(x1)) → qr(0(x1))
r1(0(x1)) → 0(r1(x1))
r0(m(x1)) → m(r0(x1))
r1(m(x1)) → m(r1(x1))
r1(1(x1)) → 1(r1(x1))
r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

r0(0(x1)) → 0(r0(x1))
r0(1(x1)) → 1(r0(x1))
r0(m(x1)) → m(r0(x1))
r1(0(x1)) → 0(r1(x1))
r1(1(x1)) → 1(r1(x1))
r1(m(x1)) → m(r1(x1))
r0(b(x1)) → qr(0(b(x1)))
r1(b(x1)) → qr(1(b(x1)))
0(qr(x1)) → qr(0(x1))
1(qr(x1)) → qr(1(x1))
m(qr(x1)) → ql(m(x1))
0(ql(x1)) → ql(0(x1))
1(ql(x1)) → ql(1(x1))
b(ql(0(x1))) → 0(b(r0(x1)))
b(ql(1(x1))) → 1(b(r1(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.