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.