Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

max1(L1(x)) -> x
max1(N2(L1(0), L1(y))) -> y
max1(N2(L1(s1(x)), L1(s1(y)))) -> s1(max1(N2(L1(x), L1(y))))
max1(N2(L1(x), N2(y, z))) -> max1(N2(L1(x), L1(max1(N2(y, z)))))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

max1(L1(x)) -> x
max1(N2(L1(0), L1(y))) -> y
max1(N2(L1(s1(x)), L1(s1(y)))) -> s1(max1(N2(L1(x), L1(y))))
max1(N2(L1(x), N2(y, z))) -> max1(N2(L1(x), L1(max1(N2(y, 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:

MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(y, z))
MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(L1(x), L1(max1(N2(y, z)))))
MAX1(N2(L1(s1(x)), L1(s1(y)))) -> MAX1(N2(L1(x), L1(y)))

The TRS R consists of the following rules:

max1(L1(x)) -> x
max1(N2(L1(0), L1(y))) -> y
max1(N2(L1(s1(x)), L1(s1(y)))) -> s1(max1(N2(L1(x), L1(y))))
max1(N2(L1(x), N2(y, z))) -> max1(N2(L1(x), L1(max1(N2(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:

MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(y, z))
MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(L1(x), L1(max1(N2(y, z)))))
MAX1(N2(L1(s1(x)), L1(s1(y)))) -> MAX1(N2(L1(x), L1(y)))

The TRS R consists of the following rules:

max1(L1(x)) -> x
max1(N2(L1(0), L1(y))) -> y
max1(N2(L1(s1(x)), L1(s1(y)))) -> s1(max1(N2(L1(x), L1(y))))
max1(N2(L1(x), N2(y, z))) -> max1(N2(L1(x), L1(max1(N2(y, z)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

MAX1(N2(L1(s1(x)), L1(s1(y)))) -> MAX1(N2(L1(x), L1(y)))

The TRS R consists of the following rules:

max1(L1(x)) -> x
max1(N2(L1(0), L1(y))) -> y
max1(N2(L1(s1(x)), L1(s1(y)))) -> s1(max1(N2(L1(x), L1(y))))
max1(N2(L1(x), N2(y, z))) -> max1(N2(L1(x), L1(max1(N2(y, z)))))

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

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

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

MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(y, z))

The TRS R consists of the following rules:

max1(L1(x)) -> x
max1(N2(L1(0), L1(y))) -> y
max1(N2(L1(s1(x)), L1(s1(y)))) -> s1(max1(N2(L1(x), L1(y))))
max1(N2(L1(x), N2(y, z))) -> max1(N2(L1(x), L1(max1(N2(y, z)))))

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