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:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(+2(x, 0)) -> F1(x)
+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:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
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:
F1(+2(x, 0)) -> F1(x)
+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:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.
↳ 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:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
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:
F1(+2(x, 0)) -> F1(x)
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
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.
F1(+2(x, 0)) -> F1(x)
Used argument filtering: F1(x1) = x1
+2(x1, x2) = +1(x1)
0 = 0
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:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
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.