Termination w.r.t. Q proof of /tmp/tmpdR7WuX/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

  ↳ 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.

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

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

M(qr(x1)) → M(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(qr(x₁)) = 1/4 + (2)x_1
POL(M(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))

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.

1¹(ql(x1)) → 1¹(x1)

The remaining pairs can at least be oriented weakly.

1¹(qr(x1)) → 1¹(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(1¹(x₁)) = (2)x_1
POL(qr(x₁)) = (2)x_1
POL(ql(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

1¹(qr(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)))

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.

1¹(qr(x1)) → 1¹(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1¹(x₁)) = (1/4)x_1
POL(qr(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))

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.

0¹(ql(x1)) → 0¹(x1)

The remaining pairs can at least be oriented weakly.

0¹(qr(x1)) → 0¹(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(0¹(x₁)) = (2)x_1
POL(ql(x₁)) = 4 + (2)x_1
POL(qr(x₁)) = (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

0¹(qr(x1)) → 0¹(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(0¹(x₁)) = (1/4)x_1
POL(qr(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 1/2 + (2)x_1
POL(m(x₁)) = 2 + (4)x_1
POL(R1(x₁)) = (1/4)x_1
POL(0(x₁)) = 1 + x_1

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x₁)) = 1/4 + x_1
POL(m(x₁)) = 1/4 + x_1
POL(R0(x₁)) = (2)x_1
POL(0(x₁)) = 2 + (4)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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.

↳ QTRS

  ↳ 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))
B(ql(0(x1))) → B(r0(x1))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(r0(x₁)) = x_1
POL(1(x₁)) = (4)x_1
POL(m(x₁)) = 4
POL(B(x₁)) = (1/4)x_1
POL(ql(x₁)) = 1/4 + (1/4)x_1
POL(qr(x₁)) = 0
POL(b(x₁)) = 0
POL(r1(x₁)) = x_1
POL(0(x₁)) = (4)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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.