Termination w.r.t. Q proof of TRS/TRCSREx2_Luc03b

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__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(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:

A__FST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
MARK₁(fst₂(X1, X2)) -> MARK₁(X2)
MARK₁(fst₂(X1, X2)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X2)
MARK₁(len₁(X)) -> MARK₁(X)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(len₁(X)) -> A__LEN₁(mark₁(X))
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(fst₂(X1, X2)) -> A__FST₂(mark₁(X1), mark₁(X2))
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(X)

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:

A__FST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
MARK₁(fst₂(X1, X2)) -> MARK₁(X2)
MARK₁(fst₂(X1, X2)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X2)
MARK₁(len₁(X)) -> MARK₁(X)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(len₁(X)) -> A__LEN₁(mark₁(X))
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(fst₂(X1, X2)) -> A__FST₂(mark₁(X1), mark₁(X2))
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

A__FST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
MARK₁(fst₂(X1, X2)) -> MARK₁(X2)
MARK₁(fst₂(X1, X2)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X2)
MARK₁(len₁(X)) -> MARK₁(X)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(fst₂(X1, X2)) -> A__FST₂(mark₁(X1), mark₁(X2))
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
A__FROM₁(X) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(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(fst₂(X1, X2)) -> MARK₁(X2)
MARK₁(fst₂(X1, X2)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X2)
MARK₁(len₁(X)) -> MARK₁(X)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(fst₂(X1, X2)) -> A__FST₂(mark₁(X1), mark₁(X2))
A__FROM₁(X) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

A__FST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__ADD₂(x₁, x₂)) = 1 + x₂
POL(A__FROM₁(x₁)) = 1 + x₁
POL(A__FST₂(x₁, x₂)) = x₂
POL(MARK₁(x₁)) = x₁
POL(a__add₂(x₁, x₂)) = 1 + x₁ + x₂
POL(a__from₁(x₁)) = 1 + x₁
POL(a__fst₂(x₁, x₂)) = 1 + x₁ + x₂
POL(a__len₁(x₁)) = 1 + x₁
POL(add₂(x₁, x₂)) = 1 + x₁ + x₂
POL(cons₂(x₁, x₂)) = x₁
POL(from₁(x₁)) = 1 + x₁
POL(fst₂(x₁, x₂)) = 1 + x₁ + x₂
POL(len₁(x₁)) = 1 + x₁
POL(mark₁(x₁)) = x₁
POL(nil) = 0
POL(s₁(x₁)) = 0

The following usable rules [14] were oriented:

a__from₁(X) -> from₁(X)
a__len₁(nil) -> 0
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(s₁(X)) -> s₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__fst₂(0, Z) -> nil
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(nil) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(len₁(X)) -> a__len₁(mark₁(X))
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
a__len₁(X) -> len₁(X)
a__fst₂(X1, X2) -> fst₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__FST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), mark₁(X2))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(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(cons₂(X1, X2)) -> MARK₁(X1)

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

POL(MARK₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), mark₁(X2))
mark₁(len₁(X)) -> a__len₁(mark₁(X))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__fst₂(X1, X2) -> fst₂(X1, X2)
a__from₁(X) -> from₁(X)
a__add₂(X1, X2) -> add₂(X1, X2)
a__len₁(X) -> len₁(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.