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:
c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(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:
C(c(c(a(x, y)))) → C(c(c(c(y))))
C(c(c(a(x, y)))) → C(y)
C(c(b(c(y), 0))) → C(a(y, 0))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
The TRS R consists of the following rules:
c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)
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:
C(c(c(a(x, y)))) → C(c(c(c(y))))
C(c(c(a(x, y)))) → C(y)
C(c(b(c(y), 0))) → C(a(y, 0))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
The TRS R consists of the following rules:
c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)
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:
C(c(c(a(x, y)))) → C(c(c(c(y))))
C(c(c(a(x, y)))) → C(y)
C(c(b(c(y), 0))) → C(a(y, 0))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(y))
C(c(a(a(y, 0), x))) → C(y)
C(c(b(c(y), 0))) → C(c(a(y, 0)))
The TRS R consists of the following rules:
c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(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
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(c(c(a(x, y)))) → C(c(c(c(y))))
C(c(c(a(x, y)))) → C(y)
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
The TRS R consists of the following rules:
c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.