(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) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(L(x1)) = x1
POL(N(x1, x2)) = 1 + x1 + x2
POL(max(x1)) = x1
POL(s(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
max(N(L(0), L(y))) → y
max(N(L(s(x)), L(s(y)))) → s(max(N(L(x), L(y))))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
max(L(x)) → x
max(N(L(x), N(y, z))) → max(N(L(x), L(max(N(y, z)))))
Q is empty.
(3) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
max(L(x)) → x
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(x0), N(x1, x2)))
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
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(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(x0), N(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(8) 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(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(x0), N(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(9) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(N(L(x), N(y, z))) → MAX(N(y, z))
R is empty.
The set Q consists of the following terms:
max(L(x0))
max(N(L(x0), N(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
(11) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
max(L(x0))
max(N(L(x0), N(x1, x2)))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(N(L(x), N(y, z))) → MAX(N(y, z))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- MAX(N(L(x), N(y, z))) → MAX(N(y, z))
The graph contains the following edges 1 > 1
(14) TRUE