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:
f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
COPY3(s1(x), y, z) -> F1(y)
COPY3(0, y, z) -> F1(z)
COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
F1(cons2(f1(cons2(nil, y)), z)) -> COPY3(n, y, z)
The TRS R consists of the following rules:
f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))
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:
COPY3(s1(x), y, z) -> F1(y)
COPY3(0, y, z) -> F1(z)
COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
F1(cons2(f1(cons2(nil, y)), z)) -> COPY3(n, y, z)
The TRS R consists of the following rules:
f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
The TRS R consists of the following rules:
f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
Used argument filtering: COPY3(x1, x2, x3) = x1
s1(x1) = s1(x1)
f1(x1) = x1
cons2(x1, x2) = x2
nil = nil
copy3(x1, x2, x3) = copy
n = n
Used ordering: Precedence:
s1 > copy
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.