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
  ↳ 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)

Q is empty.

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)

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

↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs.

↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


REM2(g2(x, y), s1(z)) -> REM2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(REM2(x1, x2)) = 3·x1 + 3·x1·x2   
POL(g2(x1, x2)) = 3 + 3·x1   
POL(s1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



↳ 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)

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.

↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(x, g2(y, z)) -> F2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x1·x2 + x2   
POL(g2(x1, x2)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ 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)

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.

↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


NORM1(g2(x, y)) -> NORM1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NORM1(x1)) = 3·x1 + 3·x12   
POL(g2(x1, x2)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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)

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.