Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
MAX(x1)  =  MAX(x1)
N(x1, x2)  =  N(x2)
L(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

[MAX₁, N₁, s₁]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
MAX(x1)  =  MAX(x1)
N(x1, x2)  =  N(x1, x2)
L(x1)  =  x1

Recursive Path Order [RPO].
Precedence:

N₂ > MAX₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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)))))

The set Q consists of the following terms:

max(L(x0))
max(N(L(0), L(x0)))
max(N(L(s(x0)), L(s(x1))))
max(N(L(x0), N(x1, x2)))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE