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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE

(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) DependencyPairsProof (EQUIVALENT transformation)

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

(2) 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)))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Complex Obligation (AND)

(5) 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)))))

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

(6) 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: Polynomial interpretation [POLO]:

POL(L(x1)) = x1   
POL(MAX(x1)) = x1   
POL(N(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) 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)))))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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)))))

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

(11) 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: Polynomial interpretation [POLO]:

POL(L(x1)) = 0   
POL(MAX(x1)) = x1   
POL(N(x1, x2)) = 1 + x2   

The following usable rules [FROCOS05] were oriented: none

(12) 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)))))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE