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:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
MINUS2(s1(x), y) -> LE2(s1(x), y)
LE2(s1(x), s1(y)) -> LE2(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)
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
MINUS2(s1(x), y) -> LE2(s1(x), y)
LE2(s1(x), s1(y)) -> LE2(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)
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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 4 SCCs with 3 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE2(s1(x), s1(y)) -> LE2(x, y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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.
LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(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
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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.
IF_MINUS3(false, s1(x), y) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2) = x1
s1(x1) = s1(x1)
IF_MINUS3(x1, x2, x3) = x2
le2(x1, x2) = le
0 = 0
false = false
true = true
Used ordering: Precedence:
le > false
le > true
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS2(s1(x), y) -> IF_MINUS3(le2(s1(x), y), s1(x), y)
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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 0 SCCs with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ 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:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
0 = 0
if_minus3(x1, x2, x3) = x2
le2(x1, x2) = le
false = false
true = true
Used ordering: Precedence:
s1 > 0
le > false
le > true
↳ QTRS
↳ Non-Overlap Check
↳ 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:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
↳ 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:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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
if_minus3(x1, x2, x3) = x2
le2(x1, x2) = le
false = false
true = true
Used ordering: Precedence:
s1 > 0
le > false
le > true
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(0, y) -> 0
minus2(s1(x), y) -> if_minus3(le2(s1(x), y), s1(x), y)
if_minus3(true, s1(x), y) -> 0
if_minus3(false, s1(x), y) -> s1(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:
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
minus2(0, x0)
minus2(s1(x0), x1)
if_minus3(true, s1(x0), x1)
if_minus3(false, s1(x0), 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.