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:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
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:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
Q DP problem:
The TRS P consists of the following rules:
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
LOG1(s1(s1(X))) -> QUOT2(X, s1(s1(0)))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))
The TRS R consists of the following rules:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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(s1(X), s1(Y)) -> MIN2(X, Y)
QUOT2(s1(X), s1(Y)) -> MIN2(X, Y)
LOG1(s1(s1(X))) -> QUOT2(X, s1(s1(0)))
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))
The TRS R consists of the following rules:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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:
MIN2(s1(X), s1(Y)) -> MIN2(X, Y)
The TRS R consists of the following rules:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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: 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:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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:
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
The TRS R consists of the following rules:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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.
QUOT2(s1(X), s1(Y)) -> QUOT2(min2(X, Y), s1(Y))
Used argument filtering: QUOT2(x1, x2) = x1
s1(x1) = s1(x1)
min2(x1, x2) = x1
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:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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:
LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))
The TRS R consists of the following rules:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
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.
LOG1(s1(s1(X))) -> LOG1(s1(quot2(X, s1(s1(0)))))
Used argument filtering: LOG1(x1) = x1
s1(x1) = s1(x1)
quot2(x1, x2) = x1
0 = 0
min2(x1, x2) = x1
Used ordering: Precedence:
trivial
↳ 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:
min2(X, 0) -> X
min2(s1(X), s1(Y)) -> min2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(min2(X, Y), s1(Y)))
log1(s1(0)) -> 0
log1(s1(s1(X))) -> s1(log1(s1(quot2(X, s1(s1(0))))))
The set Q consists of the following terms:
min2(x0, 0)
min2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
log1(s1(0))
log1(s1(s1(x0)))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.