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:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(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:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)
Q DP problem:
The TRS P consists of the following rules:
*12(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
MINUS1(p1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> MINUS1(x)
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
*12(p1(x), y) -> MINUS1(y)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
MINUS1(p1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> MINUS1(x)
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
*12(p1(x), y) -> MINUS1(y)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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:
MINUS1(s1(x)) -> MINUS1(x)
MINUS1(p1(x)) -> MINUS1(x)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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.
MINUS1(p1(x)) -> MINUS1(x)
Used argument filtering: MINUS1(x1) = x1
s1(x1) = x1
p1(x1) = p1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS1(s1(x)) -> MINUS1(x)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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.
MINUS1(s1(x)) -> MINUS1(x)
Used argument filtering: MINUS1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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)
+12(p1(x), y) -> +12(x, y)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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(p1(x), y) -> +12(x, y)
Used argument filtering: +12(x1, x2) = x1
s1(x1) = x1
p1(x1) = p1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ 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:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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(p1(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2) = x1
s1(x1) = x1
p1(x1) = p1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
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:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), 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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))
The set Q consists of the following terms:
+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.