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:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
ODD1(s1(x)) -> NOT1(odd1(x))
+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
ODD1(s1(x)) -> ODD1(x)
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
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:
ODD1(s1(x)) -> NOT1(odd1(x))
+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
ODD1(s1(x)) -> ODD1(x)
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
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: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
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(s1(x), y) -> +12(x, y)
Used argument filtering: +12(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
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
Q DP problem:
The TRS P consists of the following rules:
ODD1(s1(x)) -> ODD1(x)
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
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.
ODD1(s1(x)) -> ODD1(x)
Used argument filtering: ODD1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
not1(true) -> false
not1(false) -> true
odd1(0) -> false
odd1(s1(x)) -> not1(odd1(x))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
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.