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:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
TIMES2(s1(x), y) -> PLUS2(times2(x, y), y)
P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
TIMES2(s1(x), y) -> TIMES2(x, y)
FAC1(s1(x)) -> TIMES2(fac1(p1(s1(x))), s1(x))
PLUS2(x, s1(y)) -> PLUS2(x, y)
FAC1(s1(x)) -> FAC1(p1(s1(x)))
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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:
TIMES2(s1(x), y) -> PLUS2(times2(x, y), y)
P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
TIMES2(s1(x), y) -> TIMES2(x, y)
FAC1(s1(x)) -> TIMES2(fac1(p1(s1(x))), s1(x))
PLUS2(x, s1(y)) -> PLUS2(x, y)
FAC1(s1(x)) -> FAC1(p1(s1(x)))
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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:
P1(s1(s1(x))) -> P1(s1(x))
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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.
P1(s1(s1(x))) -> P1(s1(x))
Used argument filtering: P1(x1) = x1
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:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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:
PLUS2(x, s1(y)) -> PLUS2(x, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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.
PLUS2(x, s1(y)) -> PLUS2(x, y)
Used argument filtering: PLUS2(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:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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:
TIMES2(s1(x), y) -> TIMES2(x, y)
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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.
TIMES2(s1(x), y) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2) = x1
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:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
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:
FAC1(s1(x)) -> FAC1(p1(s1(x)))
The TRS R consists of the following rules:
plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.