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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


REM2(g2(x, y), s1(z)) -> REM2(x, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REM2(x1, x2)  =  x2
g2(x1, x2)  =  g
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(x, g2(y, z)) -> F2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x2)
g2(x1, x2)  =  g2(x1, x2)

Lexicographic Path Order [19].
Precedence:
g2 > F1


The following usable rules [14] were oriented: none



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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


NORM1(g2(x, y)) -> NORM1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
NORM1(x1)  =  NORM1(x1)
g2(x1, x2)  =  g2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[NORM1, g2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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.