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
  ↳ 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
            ↳ 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)))))

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.


MAX1(N2(L1(s1(x)), L1(s1(y)))) -> MAX1(N2(L1(x), L1(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( L1(x1) ) = x1 + 1


POL( N2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( s1(x1) ) = x1 + 1


POL( MAX1(x1) ) = max{0, x1 - 1}



The following usable rules [14] 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:

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 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:

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( N2(x1, x2) ) = x2 + 1


POL( L1(x1) ) = max{0, -1}


POL( MAX1(x1) ) = x1 + 1



The following usable rules [14] 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:

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 TRS P is empty. Hence, there is no (P,Q,R) chain.