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:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
GCD2(s1(x), s1(y)) -> -12(s1(max2(x, y)), s1(min2(x, y)))
GCD2(s1(x), s1(y)) -> GCD2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
GCD2(s1(x), s1(y)) -> MAX2(x, y)
GCD2(s1(x), s1(y)) -> MIN2(x, y)
MAX2(s1(x), s1(y)) -> MAX2(x, y)
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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:
GCD2(s1(x), s1(y)) -> -12(s1(max2(x, y)), s1(min2(x, y)))
GCD2(s1(x), s1(y)) -> GCD2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
GCD2(s1(x), s1(y)) -> MAX2(x, y)
GCD2(s1(x), s1(y)) -> MIN2(x, y)
MAX2(s1(x), s1(y)) -> MAX2(x, y)
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ 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:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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:
MAX2(s1(x), s1(y)) -> MAX2(x, y)
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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.
MAX2(s1(x), s1(y)) -> MAX2(x, y)
Used argument filtering: MAX2(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:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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:
MIN2(s1(x), s1(y)) -> MIN2(x, y)
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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:
GCD2(s1(x), s1(y)) -> GCD2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
The TRS R consists of the following rules:
min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.