Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM01

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__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(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(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__TAIL₁(cons₂(X, L)) -> MARK₁(L)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
A__NATS -> A__ADX₁(a__zeros)
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
MARK₁(head₁(X)) -> A__HEAD₁(mark₁(X))
MARK₁(zeros) -> A__ZEROS
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(head₁(X)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
MARK₁(tail₁(X)) -> MARK₁(X)
A__NATS -> A__ZEROS
A__HEAD₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(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:

MARK₁(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__TAIL₁(cons₂(X, L)) -> MARK₁(L)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
A__NATS -> A__ADX₁(a__zeros)
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
MARK₁(head₁(X)) -> A__HEAD₁(mark₁(X))
MARK₁(zeros) -> A__ZEROS
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(head₁(X)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
MARK₁(tail₁(X)) -> MARK₁(X)
A__NATS -> A__ZEROS
A__HEAD₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK₁(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__TAIL₁(cons₂(X, L)) -> MARK₁(L)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
A__NATS -> A__ADX₁(a__zeros)
MARK₁(head₁(X)) -> A__HEAD₁(mark₁(X))
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(head₁(X)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
MARK₁(tail₁(X)) -> MARK₁(X)
A__HEAD₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(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(tail₁(X)) -> A__TAIL₁(mark₁(X))
A__NATS -> A__ADX₁(a__zeros)
MARK₁(head₁(X)) -> A__HEAD₁(mark₁(X))
MARK₁(head₁(X)) -> MARK₁(X)
MARK₁(tail₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

A__TAIL₁(cons₂(X, L)) -> MARK₁(L)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> MARK₁(X)
A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
A__HEAD₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(adx₁(X)) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__ADX₁(x₁)) = x₁
POL(A__HEAD₁(x₁)) = x₁
POL(A__INCR₁(x₁)) = x₁
POL(A__NATS) = 1
POL(A__TAIL₁(x₁)) = x₁
POL(MARK₁(x₁)) = x₁
POL(a__adx₁(x₁)) = x₁
POL(a__head₁(x₁)) = 1 + x₁
POL(a__incr₁(x₁)) = x₁
POL(a__nats) = 1
POL(a__tail₁(x₁)) = 1 + x₁
POL(a__zeros) = 0
POL(adx₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(head₁(x₁)) = 1 + x₁
POL(incr₁(x₁)) = x₁
POL(mark₁(x₁)) = x₁
POL(nats) = 1
POL(nil) = 0
POL(s₁(x₁)) = x₁
POL(tail₁(x₁)) = 1 + x₁
POL(zeros) = 0

The following usable rules [14] were oriented:

a__adx₁(X) -> adx₁(X)
mark₁(0) -> 0
mark₁(nil) -> nil
a__nats -> a__adx₁(a__zeros)
mark₁(nats) -> a__nats
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
a__tail₁(X) -> tail₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__head₁(X) -> head₁(X)
mark₁(zeros) -> a__zeros
a__adx₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
a__zeros -> zeros
a__head₁(cons₂(X, L)) -> mark₁(X)
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(head₁(X)) -> a__head₁(mark₁(X))
a__tail₁(cons₂(X, L)) -> mark₁(L)
a__incr₁(nil) -> nil
a__nats -> nats
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
a__incr₁(X) -> incr₁(X)
a__zeros -> cons₂(0, zeros)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
A__TAIL₁(cons₂(X, L)) -> MARK₁(L)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
A__HEAD₁(cons₂(X, L)) -> MARK₁(X)
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__ADX₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(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:

A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__ADX₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(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__ADX₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__ADX₁(x₁)) = 1 + x₁
POL(A__INCR₁(x₁)) = x₁
POL(MARK₁(x₁)) = x₁
POL(a__adx₁(x₁)) = 1 + x₁
POL(a__head₁(x₁)) = 1 + x₁
POL(a__incr₁(x₁)) = x₁
POL(a__nats) = 1
POL(a__tail₁(x₁)) = 1 + x₁
POL(a__zeros) = 0
POL(adx₁(x₁)) = 1 + x₁
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(head₁(x₁)) = 1 + x₁
POL(incr₁(x₁)) = x₁
POL(mark₁(x₁)) = x₁
POL(nats) = 1
POL(nil) = 0
POL(s₁(x₁)) = x₁
POL(tail₁(x₁)) = 1 + x₁
POL(zeros) = 0

The following usable rules [14] were oriented:

a__adx₁(X) -> adx₁(X)
mark₁(0) -> 0
mark₁(nil) -> nil
a__nats -> a__adx₁(a__zeros)
mark₁(nats) -> a__nats
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
a__tail₁(X) -> tail₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__head₁(X) -> head₁(X)
mark₁(zeros) -> a__zeros
a__adx₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
a__zeros -> zeros
a__head₁(cons₂(X, L)) -> mark₁(X)
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(head₁(X)) -> a__head₁(mark₁(X))
a__tail₁(cons₂(X, L)) -> mark₁(L)
a__incr₁(nil) -> nil
a__nats -> nats
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
a__zeros -> cons₂(0, zeros)
a__incr₁(X) -> incr₁(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

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

A__ADX₁(cons₂(X, L)) -> A__INCR₁(cons₂(mark₁(X), adx₁(L)))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
A__INCR₁(cons₂(X, L)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__incr₁(nil) -> nil
a__incr₁(cons₂(X, L)) -> cons₂(s₁(mark₁(X)), incr₁(L))
a__adx₁(nil) -> nil
a__adx₁(cons₂(X, L)) -> a__incr₁(cons₂(mark₁(X), adx₁(L)))
a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__head₁(cons₂(X, L)) -> mark₁(X)
a__tail₁(cons₂(X, L)) -> mark₁(L)
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(nats) -> a__nats
mark₁(zeros) -> a__zeros
mark₁(head₁(X)) -> a__head₁(mark₁(X))
mark₁(tail₁(X)) -> a__tail₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(0) -> 0
a__incr₁(X) -> incr₁(X)
a__adx₁(X) -> adx₁(X)
a__nats -> nats
a__zeros -> zeros
a__head₁(X) -> head₁(X)
a__tail₁(X) -> tail₁(X)

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