Termination w.r.t. Q proof of /tmp/tmpQZNzfF/beans1.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:

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

                ↳ RuleRemovalProof

            ↳ 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.
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

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ DependencyGraphProof

            ↳ UsableRulesProof

          ↳ 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.
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

          ↳ 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.

↳ 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))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

            ↳ 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 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) ) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

            ↳ UsableRulesProof

  ↳ 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 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

  ↳ 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 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.