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:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, 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:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))
Q DP problem:
The TRS P consists of the following rules:
MIN2(x, .2(y, z)) -> MIN2(y, z)
DEL2(x, .2(y, z)) -> DEL2(x, z)
MSORT1(.2(x, y)) -> DEL2(min2(x, y), .2(x, y))
MSORT1(.2(x, y)) -> MIN2(x, y)
MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
MIN2(x, .2(y, z)) -> MIN2(x, z)
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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:
MIN2(x, .2(y, z)) -> MIN2(y, z)
DEL2(x, .2(y, z)) -> DEL2(x, z)
MSORT1(.2(x, y)) -> DEL2(min2(x, y), .2(x, y))
MSORT1(.2(x, y)) -> MIN2(x, y)
MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
MIN2(x, .2(y, z)) -> MIN2(x, z)
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
DEL2(x, .2(y, z)) -> DEL2(x, z)
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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.
DEL2(x, .2(y, z)) -> DEL2(x, z)
Used argument filtering: DEL2(x1, x2) = x2
.2(x1, x2) = .1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MIN2(x, .2(y, z)) -> MIN2(y, z)
MIN2(x, .2(y, z)) -> MIN2(x, z)
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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.
MIN2(x, .2(y, z)) -> MIN2(y, z)
MIN2(x, .2(y, z)) -> MIN2(x, z)
Used argument filtering: MIN2(x1, x2) = x2
.2(x1, x2) = .1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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.
MSORT1(.2(x, y)) -> MSORT1(del2(min2(x, y), .2(x, y)))
Used argument filtering: MSORT1(x1) = x1
.2(x1, x2) = .
del2(x1, x2) = del
min2(x1, x2) = x1
nil = nil
if3(x1, x2, x3) = if
Used ordering: Precedence:
. > del > nil > if
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
msort1(nil) -> nil
msort1(.2(x, y)) -> .2(min2(x, y), msort1(del2(min2(x, y), .2(x, y))))
min2(x, nil) -> x
min2(x, .2(y, z)) -> if3(<=2(x, y), min2(x, z), min2(y, z))
del2(x, nil) -> nil
del2(x, .2(y, z)) -> if3(=2(x, y), z, .2(y, del2(x, z)))
The set Q consists of the following terms:
msort1(nil)
msort1(.2(x0, x1))
min2(x0, nil)
min2(x0, .2(x1, x2))
del2(x0, nil)
del2(x0, .2(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.