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:
c1(c1(c1(y))) -> c1(c1(a2(y, 0)))
c1(a2(a2(0, x), y)) -> a2(c1(c1(c1(0))), y)
c1(y) -> y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c1(c1(c1(y))) -> c1(c1(a2(y, 0)))
c1(a2(a2(0, x), y)) -> a2(c1(c1(c1(0))), y)
c1(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:
C1(a2(a2(0, x), y)) -> C1(0)
C1(c1(c1(y))) -> C1(a2(y, 0))
C1(a2(a2(0, x), y)) -> C1(c1(0))
C1(a2(a2(0, x), y)) -> C1(c1(c1(0)))
C1(c1(c1(y))) -> C1(c1(a2(y, 0)))
The TRS R consists of the following rules:
c1(c1(c1(y))) -> c1(c1(a2(y, 0)))
c1(a2(a2(0, x), y)) -> a2(c1(c1(c1(0))), y)
c1(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:
C1(a2(a2(0, x), y)) -> C1(0)
C1(c1(c1(y))) -> C1(a2(y, 0))
C1(a2(a2(0, x), y)) -> C1(c1(0))
C1(a2(a2(0, x), y)) -> C1(c1(c1(0)))
C1(c1(c1(y))) -> C1(c1(a2(y, 0)))
The TRS R consists of the following rules:
c1(c1(c1(y))) -> c1(c1(a2(y, 0)))
c1(a2(a2(0, x), y)) -> a2(c1(c1(c1(0))), y)
c1(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 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C1(c1(c1(y))) -> C1(a2(y, 0))
C1(a2(a2(0, x), y)) -> C1(c1(0))
C1(a2(a2(0, x), y)) -> C1(c1(c1(0)))
C1(c1(c1(y))) -> C1(c1(a2(y, 0)))
The TRS R consists of the following rules:
c1(c1(c1(y))) -> c1(c1(a2(y, 0)))
c1(a2(a2(0, x), y)) -> a2(c1(c1(c1(0))), y)
c1(y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.