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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(x, g(y, z)) → F(x, y)
NORM(g(x, y)) → NORM(x)
REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(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:

F(x, g(y, z)) → F(x, y)
NORM(g(x, y)) → NORM(x)
REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

REM(g(x, y), s(z)) → REM(x, z)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


REM(g(x, y), s(z)) → REM(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(g(x1, x2)) = 3/4 + (9/4)x_1   
POL(s(x1)) = (13/4)x_1   
POL(REM(x1, x2)) = (3)x_1 + (15/4)x_2   
The value of delta used in the strict ordering is 9/4.
The following usable rules [17] 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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(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:

F(x, g(y, z)) → F(x, y)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, g(y, z)) → F(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(g(x1, x2)) = 1/4 + (7/2)x_1   
POL(F(x1, x2)) = (2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(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:

NORM(g(x, y)) → NORM(x)

The TRS R consists of the following rules:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


NORM(g(x, y)) → NORM(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(NORM(x1)) = (2)x_1   
POL(g(x1, x2)) = 1/4 + (7/2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(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.