Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(s1(x), s1(y)) -> TWICE1(min2(x, y))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
F2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), s1(y)) -> -12(x, min2(x, y))
F2(s1(x), s1(y)) -> F2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
F2(s1(x), s1(y)) -> F2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
TWICE1(s1(x)) -> TWICE1(x)
F2(s1(x), s1(y)) -> -12(y, min2(x, y))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F2(s1(x), s1(y)) -> TWICE1(min2(x, y))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
F2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), s1(y)) -> -12(x, min2(x, y))
F2(s1(x), s1(y)) -> F2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
F2(s1(x), s1(y)) -> F2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
TWICE1(s1(x)) -> TWICE1(x)
F2(s1(x), s1(y)) -> -12(y, min2(x, y))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TWICE1(s1(x)) -> TWICE1(x)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
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.
TWICE1(s1(x)) -> TWICE1(x)
Used argument filtering: TWICE1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MIN2(s1(x), s1(y)) -> MIN2(x, y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
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(s1(x), s1(y)) -> MIN2(x, y)
Used argument filtering: MIN2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-12(s1(x), s1(y)) -> -12(x, y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
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.
-12(s1(x), s1(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(s1(x), s1(y)) -> F2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
F2(s1(x), s1(y)) -> F2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
f2(s1(x), s1(y)) -> f2(-2(y, min2(x, y)), s1(twice1(min2(x, y))))
f2(s1(x), s1(y)) -> f2(-2(x, min2(x, y)), s1(twice1(min2(x, y))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.