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(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(y)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
+12(x, s1(y)) -> +12(x, y)
+12(x, s1(y)) -> S1(+2(x, y))
S1(+2(0, y)) -> S1(y)
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(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:
+12(x, s1(y)) -> +12(x, y)
+12(x, s1(y)) -> S1(+2(x, y))
S1(+2(0, y)) -> S1(y)
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(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:
S1(+2(0, y)) -> S1(y)
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(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.
S1(+2(0, y)) -> S1(y)
Used argument filtering: S1(x1) = x1
+2(x1, x2) = +1(x2)
0 = 0
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(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:
+12(x, s1(y)) -> +12(x, y)
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, s1(y)) -> s1(y)
s1(+2(0, y)) -> s1(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.