Termination w.r.t. Q proof of TRS/TRCSREx1_2_Luc02c

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:

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, cons₂(Y, Z))) -> MARK₁(Y)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, cons₂(Y, Z))) -> MARK₁(Y)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK₁(from₁(X)) -> MARK₁(X)
A__2ND₁(cons₂(X, cons₂(Y, Z))) -> MARK₁(Y)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(2nd₁(X)) -> A__2ND₁(mark₁(X))

The remaining pairs can at least be oriented weakly.

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)
A__FROM₁(X) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(2nd₁(x₁)) = x₁ + x₁²
POL(A__2ND₁(x₁)) = 2·x₁
POL(A__FROM₁(x₁)) = 1 + 3·x₁
POL(MARK₁(x₁)) = 1 + 2·x₁
POL(a__2nd₁(x₁)) = x₁ + x₁²
POL(a__from₁(x₁)) = 3 + x₁ + 3·x₁²
POL(cons₂(x₁, x₂)) = 1 + x₁ + x₂
POL(from₁(x₁)) = 2 + x₁ + 2·x₁²
POL(mark₁(x₁)) = x₁²
POL(s₁(x₁)) = x₁

The following usable rules [14] were oriented:

mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__from₁(X) -> from₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK₁(2nd₁(X)) -> MARK₁(X)
MARK₁(s₁(X)) -> MARK₁(X)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(2nd₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(2nd₁(x₁)) = 3 + 3·x₁ + 3·x₁²
POL(MARK₁(x₁)) = 3·x₁ + 3·x₁²
POL(s₁(x₁)) = 3 + x₁ + 2·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

a__2nd₁(cons₂(X, cons₂(Y, Z))) -> mark₁(Y)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(2nd₁(X)) -> a__2nd₁(mark₁(X))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__2nd₁(X) -> 2nd₁(X)
a__from₁(X) -> from₁(X)

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.