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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))


Q DP problem:
The TRS P consists of the following rules:

G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> G1(f1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
G1(s1(x)) -> MINUS2(s1(x), f1(g1(x)))
F1(s1(x)) -> MINUS2(s1(x), g1(f1(x)))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> G1(f1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
G1(s1(x)) -> MINUS2(s1(x), f1(g1(x)))
F1(s1(x)) -> MINUS2(s1(x), g1(f1(x)))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

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.

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

Q DP problem:
The TRS P consists of the following rules:

G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> G1(f1(x))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

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.

G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> F1(x)
Used argument filtering: G1(x1)  =  x1
s1(x1)  =  s1(x1)
F1(x1)  =  x1
g1(x1)  =  x1
f1(x1)  =  f1(x1)
0  =  0
minus2(x1, x2)  =  x1
Used ordering: Quasi Precedence: [s_1, f_1] 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F1(s1(x)) -> G1(f1(x))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.