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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
MINUS2(x, s1(y)) -> PRED1(minus2(x, y))
LOG1(s1(s1(x))) -> QUOT2(x, s1(s1(0)))
LOG1(s1(s1(x))) -> LOG1(s1(quot2(x, s1(s1(0)))))
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(x, s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
MINUS2(x, s1(y)) -> PRED1(minus2(x, y))
LOG1(s1(s1(x))) -> QUOT2(x, s1(s1(0)))
LOG1(s1(s1(x))) -> LOG1(s1(quot2(x, s1(s1(0)))))
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(x, s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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 3 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:
MINUS2(x, s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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.
MINUS2(x, s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi 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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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(minus2(x, y), s1(y))
The TRS R consists of the following rules:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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(minus2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2) = x1
s1(x1) = s1(x1)
minus2(x1, x2) = x1
pred1(x1) = x1
Used ordering: Quasi 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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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
minus2(x1, x2) = x1
pred1(x1) = x1
Used ordering: Quasi 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:
pred1(s1(x)) -> x
minus2(x, 0) -> x
minus2(x, s1(y)) -> pred1(minus2(x, y))
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(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:
pred1(s1(x0))
minus2(x0, 0)
minus2(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.