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:
minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MINUS1(h1(x)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)
The TRS R consists of the following rules:
minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MINUS1(h1(x)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)
The TRS R consists of the following rules:
minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))
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.
MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)
Used argument filtering: MINUS1(x1) = x1
h1(x1) = x1
f2(x1, x2) = f2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MINUS1(h1(x)) -> MINUS1(x)
The TRS R consists of the following rules:
minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))
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.
MINUS1(h1(x)) -> MINUS1(x)
Used argument filtering: MINUS1(x1) = x1
h1(x1) = h1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))
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.