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:

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
  ↳ Non-Overlap Check

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.

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:

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)))))

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, 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:

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)))))

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, 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:

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)))))

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, x2)))

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

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, 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.


MAX1(N2(L1(s1(x)), L1(s1(y)))) -> MAX1(N2(L1(x), L1(y)))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MAX1(x1)  =  MAX1(x1)
N2(x1, x2)  =  N2(x1, x2)
L1(x1)  =  x1
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
MAX1 > [N2, s1]


The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
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)))))

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, 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

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)))))

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, 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.


MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(y, z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MAX1(x1)  =  MAX1(x1)
N2(x1, x2)  =  N2(x1, x2)
L1(x1)  =  L

Lexicographic Path Order [19].
Precedence:
[MAX1, N2]


The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
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)))))

The set Q consists of the following terms:

max1(L1(x0))
max1(N2(L1(0), L1(x0)))
max1(N2(L1(s1(x0)), L1(s1(x1))))
max1(N2(L1(x0), N2(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.