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)))

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
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)

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:

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)))

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
          ↳ 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)

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
          ↳ 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)

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:

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)))

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
          ↳ 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)

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
          ↳ 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)

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:

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)))

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
          ↳ 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)

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
          ↳ 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)

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:

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.