(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
The set Q consists of the following terms:
r(x0, x1, x2, nil)
r(x0, nil, x1, cons(x2, x3))
r(x0, cons(x1, x2), nil, cons(x3, x4))
r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), ws)
R(xs, cons(y, ys), nil, cons(w, ws)) → R(xs, xs, cons(succ(zero), nil), ws)
R(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → R(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
The set Q consists of the following terms:
r(x0, x1, x2, nil)
r(x0, nil, x1, cons(x2, x3))
r(x0, cons(x1, x2), nil, cons(x3, x4))
r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6))
We have to consider all minimal (P,Q,R)-chains.