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)))
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 contains 2 SCCs with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MAX1(N2(L1(s1(x)), L1(s1(y)))) -> MAX1(N2(L1(x), L1(y)))
Used argument filtering: MAX1(x1) = x1
N2(x1, x2) = x2
L1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ 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
↳ QDPAfsSolverProof
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MAX1(N2(L1(x), N2(y, z))) -> MAX1(N2(y, z))
Used argument filtering: MAX1(x1) = x1
N2(x1, x2) = N1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ 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.