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(*2(x, y), z) -> *2(x, *2(y, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, +2(y, f1(z))) -> *2(g2(x, z), +2(y, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, +2(y, f1(z))) -> *2(g2(x, z), +2(y, y))
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:
*12(+2(x, y), z) -> *12(x, z)
*12(*2(x, y), z) -> *12(y, z)
*12(+2(x, y), z) -> *12(y, z)
*12(x, +2(y, f1(z))) -> *12(g2(x, z), +2(y, y))
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, +2(y, f1(z))) -> *2(g2(x, z), +2(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(+2(x, y), z) -> *12(x, z)
*12(*2(x, y), z) -> *12(y, z)
*12(+2(x, y), z) -> *12(y, z)
*12(x, +2(y, f1(z))) -> *12(g2(x, z), +2(y, y))
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, +2(y, f1(z))) -> *2(g2(x, z), +2(y, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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, f1(z))) -> *12(g2(x, z), +2(y, y))
The TRS R consists of the following rules:
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, +2(y, f1(z))) -> *2(g2(x, z), +2(y, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
*12(+2(x, y), z) -> *12(x, z)
*12(*2(x, y), z) -> *12(y, z)
*12(+2(x, y), z) -> *12(y, z)
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
*2(*2(x, y), z) -> *2(x, *2(y, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, +2(y, f1(z))) -> *2(g2(x, z), +2(y, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.