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:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), 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(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → A(c(c(c(a(x, 0)))), y)
C(a(c(c(y)), x)) → A(x, 0)
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(a(x, 0))
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), 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(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → A(c(c(c(a(x, 0)))), y)
C(a(c(c(y)), x)) → A(x, 0)
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(a(x, 0))
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), 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(a(c(c(y)), x)) → A(c(c(c(a(x, 0)))), y)
C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → A(x, 0)
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(a(x, 0))
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(a(x, 0))
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.