Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM01_GM.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(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:

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, L)) → MARK(L)
MARK(adx(X)) → A__ADX(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(incr(X)) → A__INCR(mark(X))
MARK(head(X)) → A__HEAD(mark(X))
MARK(zeros) → A__ZEROS
A__INCR(cons(X, L)) → MARK(X)
MARK(head(X)) → MARK(X)
A__ADX(cons(X, L)) → MARK(X)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(s(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(tail(X)) → MARK(X)
A__NATS → A__ZEROS
A__HEAD(cons(X, L)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(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:

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, L)) → MARK(L)
MARK(adx(X)) → A__ADX(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(incr(X)) → A__INCR(mark(X))
MARK(head(X)) → A__HEAD(mark(X))
MARK(zeros) → A__ZEROS
A__INCR(cons(X, L)) → MARK(X)
MARK(head(X)) → MARK(X)
A__ADX(cons(X, L)) → MARK(X)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(s(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(tail(X)) → MARK(X)
A__NATS → A__ZEROS
A__HEAD(cons(X, L)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(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:

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, L)) → MARK(L)
MARK(adx(X)) → A__ADX(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(zeros) → A__ZEROS
MARK(head(X)) → A__HEAD(mark(X))
A__INCR(cons(X, L)) → MARK(X)
MARK(head(X)) → MARK(X)
A__ADX(cons(X, L)) → MARK(X)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(s(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(tail(X)) → MARK(X)
A__NATS → A__ZEROS
A__HEAD(cons(X, L)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(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

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

MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, L)) → MARK(L)
MARK(adx(X)) → A__ADX(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(head(X)) → A__HEAD(mark(X))
A__INCR(cons(X, L)) → MARK(X)
MARK(head(X)) → MARK(X)
A__ADX(cons(X, L)) → MARK(X)
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(s(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(tail(X)) → MARK(X)
A__HEAD(cons(X, L)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

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