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