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:
+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> Y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
+12(X, s1(Y)) -> +12(X, Y)
F3(0, s1(0), X) -> DOUBLE1(X)
F3(0, s1(0), X) -> F3(X, double1(X), X)
DOUBLE1(X) -> +12(X, X)
The TRS R consists of the following rules:
+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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)
F3(0, s1(0), X) -> DOUBLE1(X)
F3(0, s1(0), X) -> F3(X, double1(X), X)
DOUBLE1(X) -> +12(X, X)
The TRS R consists of the following rules:
+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> Y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.
↳ 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)
The TRS R consists of the following rules:
+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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: 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))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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
Q DP problem:
The TRS P consists of the following rules:
F3(0, s1(0), X) -> F3(X, double1(X), X)
The TRS R consists of the following rules:
+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> Y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.