Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHEx2_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 strictly oriented and are deleted.

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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
A__FST₂(x1, x2) = A__FST₂(x1, x2)
s₁(x1) = s
cons₂(x1, x2) = cons₁(x1)
MARK₁(x1) = x1
fst₂(x1, x2) = fst₂(x1, x2)
add₂(x1, x2) = add₂(x1, x2)
len₁(x1) = len₁(x1)
A__ADD₂(x1, x2) = A__ADD₁(x2)
0 = 0
from₁(x1) = from₁(x1)
A__FROM₁(x1) = A__FROM₁(x1)
mark₁(x1) = x1
a__fst₂(x1, x2) = a__fst₂(x1, x2)
a__from₁(x1) = a__from₁(x1)
a__add₂(x1, x2) = a__add₂(x1, x2)
a__len₁(x1) = a__len₁(x1)
nil = nil

Lexicographic Path Order [19].
Precedence:

[s, fst₂, add₂, len₁, 0, a_{_fst₂}, a_{_add₂}, a_{_len₁}, nil] > A_{_FST₂}
[s, fst₂, add₂, len₁, 0, a_{_fst₂}, a_{_add₂}, a_{_len₁}, nil] > [cons₁, from₁, a_{_{from_1]}} > A_{_FROM₁}
[s, fst₂, add₂, len₁, 0, a_{_fst₂}, a_{_add₂}, a_{_len₁}, nil] > A_{_ADD₁}

The following usable rules [14] were oriented:

mark₁(fst₂(X1, X2)) -> a__fst₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
a__add₂(0, X) -> 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__len₁(nil) -> 0
a__len₁(cons₂(X, Z)) -> s₁(len₁(Z))
a__len₁(X) -> len₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__add₂(X1, X2) -> add₂(X1, X2)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__from₁(X) -> from₁(X)
a__fst₂(0, Z) -> nil
a__fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), fst₂(X, Z))
a__fst₂(X1, X2) -> fst₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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.