Termination w.r.t. Q proof of TRS/TRCSREx4_7_77_Bor03

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:

zeros -> cons₂(0, n__zeros)
tail₁(cons₂(X, XS)) -> activate₁(XS)
zeros -> n__zeros
activate₁(n__zeros) -> zeros
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

zeros -> cons₂(0, n__zeros)
tail₁(cons₂(X, XS)) -> activate₁(XS)
zeros -> n__zeros
activate₁(n__zeros) -> zeros
activate₁(X) -> 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:

ACTIVATE₁(n__zeros) -> ZEROS
TAIL₁(cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

zeros -> cons₂(0, n__zeros)
tail₁(cons₂(X, XS)) -> activate₁(XS)
zeros -> n__zeros
activate₁(n__zeros) -> zeros
activate₁(X) -> 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:

ACTIVATE₁(n__zeros) -> ZEROS
TAIL₁(cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

zeros -> cons₂(0, n__zeros)
tail₁(cons₂(X, XS)) -> activate₁(XS)
zeros -> n__zeros
activate₁(n__zeros) -> zeros
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.