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:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, 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:

MAX(N(L(x), N(y, z))) → MAX(N(y, z))
MAX(N(L(x), N(y, z))) → MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, 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:

MAX(N(L(x), N(y, z))) → MAX(N(y, z))
MAX(N(L(x), N(y, z))) → MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

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

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

MAX(N(L(s(x)), L(s(y)))) → MAX(N(L(x), L(y)))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, 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.


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

POL(N(x1, x2)) = (2)x_2   
POL(L(x1)) = (4)x_1   
POL(s(x1)) = 1/2 + x_1   
POL(MAX(x1)) = x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



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

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, 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

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

MAX(N(L(x), N(y, z))) → MAX(N(y, z))

The TRS R consists of the following rules:

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, 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.


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

POL(N(x1, x2)) = 2 + (4)x_1 + (2)x_2   
POL(L(x1)) = 0   
POL(MAX(x1)) = (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



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

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

max(L(x)) → x
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, 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.