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:
div2(X, e) -> i1(X)
i1(div2(X, Y)) -> div2(Y, X)
div2(div2(X, Y), Z) -> div2(Y, div2(i1(X), Z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
div2(X, e) -> i1(X)
i1(div2(X, Y)) -> div2(Y, X)
div2(div2(X, Y), Z) -> div2(Y, div2(i1(X), Z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
I1(div2(X, Y)) -> DIV2(Y, X)
DIV2(X, e) -> I1(X)
DIV2(div2(X, Y), Z) -> I1(X)
DIV2(div2(X, Y), Z) -> DIV2(i1(X), Z)
DIV2(div2(X, Y), Z) -> DIV2(Y, div2(i1(X), Z))
The TRS R consists of the following rules:
div2(X, e) -> i1(X)
i1(div2(X, Y)) -> div2(Y, X)
div2(div2(X, Y), Z) -> div2(Y, div2(i1(X), Z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
I1(div2(X, Y)) -> DIV2(Y, X)
DIV2(X, e) -> I1(X)
DIV2(div2(X, Y), Z) -> I1(X)
DIV2(div2(X, Y), Z) -> DIV2(i1(X), Z)
DIV2(div2(X, Y), Z) -> DIV2(Y, div2(i1(X), Z))
The TRS R consists of the following rules:
div2(X, e) -> i1(X)
i1(div2(X, Y)) -> div2(Y, X)
div2(div2(X, Y), Z) -> div2(Y, div2(i1(X), Z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.