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:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

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:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))


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

NORM1(g2(x, y)) -> NORM1(x)
F2(x, g2(y, z)) -> F2(x, y)
REM2(g2(x, y), s1(z)) -> REM2(x, z)

The TRS R consists of the following rules:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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:

NORM1(g2(x, y)) -> NORM1(x)
F2(x, g2(y, z)) -> F2(x, y)
REM2(g2(x, y), s1(z)) -> REM2(x, z)

The TRS R consists of the following rules:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs.

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

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

REM2(g2(x, y), s1(z)) -> REM2(x, z)

The TRS R consists of the following rules:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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.

REM2(g2(x, y), s1(z)) -> REM2(x, z)
Used argument filtering: REM2(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
              ↳ QDP

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

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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
              ↳ QDP

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

F2(x, g2(y, z)) -> F2(x, y)

The TRS R consists of the following rules:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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.

F2(x, g2(y, z)) -> F2(x, y)
Used argument filtering: F2(x1, x2)  =  x2
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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
              ↳ QDP
                ↳ QDPAfsSolverProof

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

NORM1(g2(x, y)) -> NORM1(x)

The TRS R consists of the following rules:

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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.

NORM1(g2(x, y)) -> NORM1(x)
Used argument filtering: NORM1(x1)  =  x1
g2(x1, x2)  =  g1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)

The set Q consists of the following terms:

norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))

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