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:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(x1))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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:

01(2(0(x1))) → 11(x1)
01(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
L1(2(1(x1))) → 01(2(x1))
01(2(R(x1))) → 11(0(1(R(x1))))
11(2(1(x1))) → 01(2(x1))
L1(2(1(x1))) → L1(1(0(2(x1))))
01(2(0(x1))) → 11(0(1(x1)))
01(2(0(x1))) → 01(1(x1))
01(2(R(x1))) → 01(1(R(x1)))
L1(2(0(x1))) → 11(0(1(x1)))
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
L1(2(1(x1))) → 11(0(2(x1)))
11(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → 01(1(x1))
11(2(R(x1))) → 01(1(R(x1)))
01(2(1(x1))) → 01(2(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(x1))))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

01(2(0(x1))) → 11(x1)
01(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
L1(2(1(x1))) → 01(2(x1))
01(2(R(x1))) → 11(0(1(R(x1))))
11(2(1(x1))) → 01(2(x1))
L1(2(1(x1))) → L1(1(0(2(x1))))
01(2(0(x1))) → 11(0(1(x1)))
01(2(0(x1))) → 01(1(x1))
01(2(R(x1))) → 01(1(R(x1)))
L1(2(0(x1))) → 11(0(1(x1)))
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
L1(2(1(x1))) → 11(0(2(x1)))
11(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → 01(1(x1))
11(2(R(x1))) → 01(1(R(x1)))
01(2(1(x1))) → 01(2(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
          ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

01(2(0(x1))) → 11(x1)
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
11(2(1(x1))) → 01(2(x1))
01(2(0(x1))) → 01(1(x1))
01(2(0(x1))) → 11(0(1(x1)))
01(2(1(x1))) → 01(2(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

01(2(0(x1))) → 11(x1)
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
11(2(1(x1))) → 01(2(x1))
01(2(1(x1))) → 01(2(x1))
01(2(0(x1))) → 11(0(1(x1)))
01(2(0(x1))) → 01(1(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
          ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

01(2(0(x1))) → 11(x1)
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
11(2(1(x1))) → 01(2(x1))
01(2(1(x1))) → 01(2(x1))
01(2(0(x1))) → 11(0(1(x1)))
01(2(0(x1))) → 01(1(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

01(2(0(x1))) → 11(x1)
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
11(2(1(x1))) → 01(2(x1))
01(2(1(x1))) → 01(2(x1))
01(2(0(x1))) → 01(1(x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x1)) = x1   
POL(01(x1)) = 1 + x1   
POL(1(x1)) = 1 + x1   
POL(11(x1)) = 2 + x1   
POL(2(x1)) = 2 + x1   
POL(R(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
                  ↳ QDP
                    ↳ DependencyGraphProof
          ↳ QDP
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

01(2(1(x1))) → 11(0(2(x1)))
01(2(0(x1))) → 11(0(1(x1)))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(1(x1))) → L1(1(0(2(x1))))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
            ↳ UsableRulesProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(1(x1))) → L1(1(0(2(x1))))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(1(x1))) → L1(1(0(2(x1))))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(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.


L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(1(x1))) → L1(1(0(2(x1))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( 1(x1) ) = x1


POL( L1(x1) ) = x1


POL( 0(x1) ) = max{0, -1}


POL( 2(x1) ) = 1


POL( R(x1) ) = x1 + 1



The following usable rules [17] were oriented:

1(2(1(x1))) → 2(0(2(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
1(2(0(x1))) → 2(0(1(x1)))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(x1))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

1(2(1(x1))) → 2(0(2(x1)))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(1(x1))) → 1(0(2(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(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.
We have reversed the following QTRS:
The set of rules R is

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(x1))))

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

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(L(x))) → 2(0(1(L(x))))
0(2(1(x))) → 1(0(2(x)))
R(2(1(x))) → R(1(0(2(x))))
0(2(0(x))) → 1(0(1(x)))
0(2(L(x))) → 1(0(1(L(x))))
R(2(0(x))) → R(1(0(1(x))))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse

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

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(L(x))) → 2(0(1(L(x))))
0(2(1(x))) → 1(0(2(x)))
R(2(1(x))) → R(1(0(2(x))))
0(2(0(x))) → 1(0(1(x)))
0(2(L(x))) → 1(0(1(L(x))))
R(2(0(x))) → R(1(0(1(x))))

Q is empty.

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

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(x1))))

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

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(L(x))) → 2(0(1(L(x))))
0(2(1(x))) → 1(0(2(x)))
R(2(1(x))) → R(1(0(2(x))))
0(2(0(x))) → 1(0(1(x)))
0(2(L(x))) → 1(0(1(L(x))))
R(2(0(x))) → R(1(0(1(x))))

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS

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

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(L(x))) → 2(0(1(L(x))))
0(2(1(x))) → 1(0(2(x)))
R(2(1(x))) → R(1(0(2(x))))
0(2(0(x))) → 1(0(1(x)))
0(2(L(x))) → 1(0(1(L(x))))
R(2(0(x))) → R(1(0(1(x))))

Q is empty.