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:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
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:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
Q DP problem:
The TRS P consists of the following rules:
*12(s1(x), y) -> +12(*2(x, y), y)
+12(x, s1(y)) -> +12(x, y)
*12(s1(x), y) -> *12(x, y)
FACT1(s1(x)) -> P1(s1(x))
FACT1(s1(x)) -> *12(s1(x), fact1(p1(s1(x))))
FACT1(s1(x)) -> FACT1(p1(s1(x)))
The TRS R consists of the following rules:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
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:
*12(s1(x), y) -> +12(*2(x, y), y)
+12(x, s1(y)) -> +12(x, y)
*12(s1(x), y) -> *12(x, y)
FACT1(s1(x)) -> P1(s1(x))
FACT1(s1(x)) -> *12(s1(x), fact1(p1(s1(x))))
FACT1(s1(x)) -> FACT1(p1(s1(x)))
The TRS R consists of the following rules:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
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:
+12(x, s1(y)) -> +12(x, y)
The TRS R consists of the following rules:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
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(x, s1(y)) -> +12(x, y)
Used argument filtering: +12(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:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
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:
*12(s1(x), y) -> *12(x, y)
The TRS R consists of the following rules:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
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), y) -> *12(x, y)
Used argument filtering: *12(x1, x2) = x1
s1(x1) = s1(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:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
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
Q DP problem:
The TRS P consists of the following rules:
FACT1(s1(x)) -> FACT1(p1(s1(x)))
The TRS R consists of the following rules:
p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
The set Q consists of the following terms:
p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))
We have to consider all minimal (P,Q,R)-chains.