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:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
+12(x, +2(y, z)) -> +12(x, y)
I1(+2(x, y)) -> I1(x)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
I1(+2(x, y)) -> I1(y)
I1(+2(x, y)) -> +12(i1(x), i1(y))
The TRS R consists of the following rules:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> 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:
+12(x, +2(y, z)) -> +12(x, y)
I1(+2(x, y)) -> I1(x)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
I1(+2(x, y)) -> I1(y)
I1(+2(x, y)) -> +12(i1(x), i1(y))
The TRS R consists of the following rules:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
I1(+2(x, y)) -> I1(x)
I1(+2(x, y)) -> I1(y)
The TRS R consists of the following rules:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> 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.
I1(+2(x, y)) -> I1(x)
I1(+2(x, y)) -> I1(y)
Used argument filtering: I1(x1) = x1
+2(x1, x2) = +2(x1, x2)
Used ordering: 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:
i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> 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.