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