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(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 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(b(y, c(x))) → C(b(a(0, 0), y))
A(x, y) → B(0, c(y))
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → A(0, 0)
C(b(y, c(x))) → B(a(0, 0), y)
A(x, y) → C(y)
A(x, y) → B(x, b(0, c(y)))
The TRS R consists of the following rules:
a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 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(b(y, c(x))) → C(b(a(0, 0), y))
A(x, y) → B(0, c(y))
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → A(0, 0)
C(b(y, c(x))) → B(a(0, 0), y)
A(x, y) → C(y)
A(x, y) → B(x, b(0, c(y)))
The TRS R consists of the following rules:
a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 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:
A(x, y) → B(0, c(y))
C(b(y, c(x))) → C(b(a(0, 0), y))
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → B(a(0, 0), y)
C(b(y, c(x))) → A(0, 0)
A(x, y) → C(y)
A(x, y) → B(x, b(0, c(y)))
The TRS R consists of the following rules:
a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(b(y, c(x))) → C(b(a(0, 0), y))
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → A(0, 0)
A(x, y) → C(y)
The TRS R consists of the following rules:
a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.