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:
merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(++2(x, y), v)
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))
The TRS R consists of the following rules:
merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))
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:
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(++2(x, y), v)
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))
The TRS R consists of the following rules:
merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))
The TRS R consists of the following rules:
merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))
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.
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))
Used argument filtering: MERGE2(x1, x2) = x1
++2(x1, x2) = ++1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))
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.