Termination w.r.t. Q proof of TRS/ThiemannsizeChange.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:

r₄(xs, ys, zs, nil) -> xs
r₄(xs, nil, zs, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), zs), ws)
r₄(xs, cons₂(y, ys), nil, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), nil), ws)
r₄(xs, cons₂(y, ys), cons₂(z, zs), cons₂(w, ws)) -> r₄(ys, cons₂(y, ys), zs, cons₂(succ₁(zero), cons₂(w, ws)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

r₄(xs, ys, zs, nil) -> xs
r₄(xs, nil, zs, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), zs), ws)
r₄(xs, cons₂(y, ys), nil, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), nil), ws)
r₄(xs, cons₂(y, ys), cons₂(z, zs), cons₂(w, ws)) -> r₄(ys, cons₂(y, ys), zs, cons₂(succ₁(zero), cons₂(w, ws)))

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:

R₄(xs, cons₂(y, ys), nil, cons₂(w, ws)) -> R₄(xs, xs, cons₂(succ₁(zero), nil), ws)
R₄(xs, cons₂(y, ys), cons₂(z, zs), cons₂(w, ws)) -> R₄(ys, cons₂(y, ys), zs, cons₂(succ₁(zero), cons₂(w, ws)))
R₄(xs, nil, zs, cons₂(w, ws)) -> R₄(xs, xs, cons₂(succ₁(zero), zs), ws)

The TRS R consists of the following rules:

r₄(xs, ys, zs, nil) -> xs
r₄(xs, nil, zs, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), zs), ws)
r₄(xs, cons₂(y, ys), nil, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), nil), ws)
r₄(xs, cons₂(y, ys), cons₂(z, zs), cons₂(w, ws)) -> r₄(ys, cons₂(y, ys), zs, cons₂(succ₁(zero), cons₂(w, ws)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

R₄(xs, cons₂(y, ys), nil, cons₂(w, ws)) -> R₄(xs, xs, cons₂(succ₁(zero), nil), ws)
R₄(xs, cons₂(y, ys), cons₂(z, zs), cons₂(w, ws)) -> R₄(ys, cons₂(y, ys), zs, cons₂(succ₁(zero), cons₂(w, ws)))
R₄(xs, nil, zs, cons₂(w, ws)) -> R₄(xs, xs, cons₂(succ₁(zero), zs), ws)

The TRS R consists of the following rules:

r₄(xs, ys, zs, nil) -> xs
r₄(xs, nil, zs, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), zs), ws)
r₄(xs, cons₂(y, ys), nil, cons₂(w, ws)) -> r₄(xs, xs, cons₂(succ₁(zero), nil), ws)
r₄(xs, cons₂(y, ys), cons₂(z, zs), cons₂(w, ws)) -> r₄(ys, cons₂(y, ys), zs, cons₂(succ₁(zero), cons₂(w, ws)))

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