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:
g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G(X) → H(X)
F(k(a), k(b), X) → F(X, X, X)
G(X) → U(h(X), h(X), X)
The TRS R consists of the following rules:
g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
G(X) → H(X)
F(k(a), k(b), X) → F(X, X, X)
G(X) → U(h(X), h(X), X)
The TRS R consists of the following rules:
g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G(X) → H(X)
F(k(a), k(b), X) → F(X, X, X)
G(X) → U(h(X), h(X), X)
The TRS R consists of the following rules:
g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(k(a), k(b), X) → F(X, X, X)
The TRS R consists of the following rules:
g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.