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__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(len(X)) → A__LEN(mark(X))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(len(X)) → A__LEN(mark(X))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(len(X)) → A__LEN(mark(X))
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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.

A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X2)
MARK(fst(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(from(X)) → MARK(X)
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__FROM(X) → MARK(X)

The remaining pairs can at least be oriented weakly.

A__ADD(0, X) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)

Used ordering: Combined order from the following AFS and order.
A__FST(x1, x2) = A__FST(x2)
s(x1) = s
cons(x1, x2) = x1
MARK(x1) = x1
fst(x1, x2) = fst(x1, x2)
add(x1, x2) = add(x1, x2)
len(x1) = len(x1)
A__ADD(x1, x2) = x2
0 = 0
from(x1) = from(x1)
A__FROM(x1) = A__FROM(x1)
mark(x1) = x1
a__add(x1, x2) = a__add(x1, x2)
a__len(x1) = a__len(x1)
nil = nil
a__from(x1) = a__from(x1)
a__fst(x1, x2) = a__fst(x1, x2)

Lexicographic Path Order [19].
Precedence:

[s, len₁, 0, from₁, A_{_FROM₁}, a_{_len₁}, a_{_{from_1]}} > [fst₂, a_{_{fst_2]}} > A_{_FST₁}
[s, len₁, 0, from₁, A_{_FROM₁}, a_{_len₁}, a_{_{from_1]}} > [fst₂, a_{_{fst_2]}} > nil
[s, len₁, 0, from₁, A_{_FROM₁}, a_{_len₁}, a_{_{from_1]}} > [add₂, a_{_{add_2]}}

The following usable rules [14] were oriented:

mark(cons(X1, X2)) → cons(mark(X1), X2)
a__add(0, X) → mark(X)
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__len(nil) → 0
a__from(X) → from(X)
mark(nil) → nil
mark(len(X)) → a__len(mark(X))
a__from(X) → cons(mark(X), from(s(X)))
a__len(cons(X, Z)) → s(len(Z))
mark(s(X)) → s(X)
mark(0) → 0
a__len(X) → len(X)
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
a__fst(X1, X2) → fst(X1, X2)
mark(from(X)) → a__from(mark(X))
a__add(s(X), Y) → s(add(X, Y))
a__fst(0, Z) → nil
a__add(X1, X2) → add(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

A__ADD(0, X) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
cons(x1, x2) = cons(x1)

Lexicographic Path Order [19].
Precedence:

[MARK₁, cons_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(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.