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:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
*12(x, s1(y)) -> +12(*2(x, y), x)
FLOOP2(s1(x), y) -> *12(s1(x), y)
FAC1(s1(x)) -> FAC1(x)
+12(x, s1(y)) -> +12(x, y)
FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
FAC1(s1(x)) -> *12(s1(x), fac1(x))
*12(x, s1(y)) -> *12(x, y)
FAC1(0) -> 11
The TRS R consists of the following rules:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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:
*12(x, s1(y)) -> +12(*2(x, y), x)
FLOOP2(s1(x), y) -> *12(s1(x), y)
FAC1(s1(x)) -> FAC1(x)
+12(x, s1(y)) -> +12(x, y)
FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
FAC1(s1(x)) -> *12(s1(x), fac1(x))
*12(x, s1(y)) -> *12(x, y)
FAC1(0) -> 11
The TRS R consists of the following rules:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ 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:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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.
+12(x, s1(y)) -> +12(x, y)
Used argument filtering: +12(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: 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:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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:
*12(x, s1(y)) -> *12(x, y)
The TRS R consists of the following rules:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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.
*12(x, s1(y)) -> *12(x, y)
Used argument filtering: *12(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: 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:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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:
FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
The TRS R consists of the following rules:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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.
FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
Used argument filtering: FLOOP2(x1, x2) = x1
s1(x1) = s1(x1)
*2(x1, x2) = *2(x1, x2)
0 = 0
+2(x1, x2) = +2(x1, x2)
Used ordering: Precedence:
*2 > +2 > s1
↳ 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:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
FAC1(s1(x)) -> FAC1(x)
The TRS R consists of the following rules:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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.
FAC1(s1(x)) -> FAC1(x)
Used argument filtering: FAC1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ 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:
fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)
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.