let R be the TRS under consideration

a(a(_1)) -> b(a(b(_1))) is in elim_R(R)
let l0 be the left-hand side of this rule
p0 = 0 is a position in l0
we have l0|p0 = a(_1)
b(b(b(_2))) -> a(a(_2)) is in R
let r'0 be the right-hand side of this rule
theta0 = {_1/a(_2)} is a mgu of l0|p0 and r'0

==> a(b(b(b(_1)))) -> b(a(b(a(_1)))) is in EU_R^1
let l1 be the left-hand side of this rule
p1 = 0.0.0 is a position in l1
we have l1|p1 = b(_1)
a(a(_2)) -> b(a(b(_2))) is in R
let r'1 be the right-hand side of this rule
theta1 = {_1/a(b(_2))} is a mgu of l1|p1 and r'1

==> a(b(b(a(a(_1))))) -> b(a(b(a(a(b(_1)))))) is in EU_R^2
let l2 be the left-hand side of this rule
p2 = 0.0.0.0 is a position in l2
we have l2|p2 = a(_1)
b(b(b(_2))) -> a(a(_2)) is in R
let r'2 be the right-hand side of this rule
theta2 = {_1/a(_2)} is a mgu of l2|p2 and r'2

==> a(b(b(a(b(b(b(_1))))))) -> b(a(b(a(a(b(a(_1))))))) is in EU_R^3
let l3 be the left-hand side of this rule
p3 = 0.0 is a position in l3
we have l3|p3 = b(a(b(b(b(_1)))))
a(a(_2)) -> b(a(b(_2))) is in R
let r'3 be the right-hand side of this rule
theta3 = {_2/b(b(_1))} is a mgu of l3|p3 and r'3

==> a(b(a(a(b(b(_1)))))) -> b(a(b(a(a(b(a(_1))))))) is in EU_R^4
let l4 be the left-hand side of this rule
p4 = 0.0.0.0.0 is a position in l4
we have l4|p4 = b(_1)
a(a(_2)) -> b(a(b(_2))) is in R
let r'4 be the right-hand side of this rule
theta4 = {_1/a(b(_2))} is a mgu of l4|p4 and r'4

==> a(b(a(a(b(a(a(_1))))))) -> b(a(b(a(a(b(a(a(b(_1))))))))) is in EU_R^5
let l be the left-hand side and r be the right-hand side of this rule
let p = 0
let theta = {}
let theta' = {_1/b(_1)}
we have r|p = a(b(a(a(b(a(a(b(_1)))))))) and
theta'(theta(l)) = theta(r|p)
so, theta(l) = a(b(a(a(b(a(a(_1))))))) is non-terminating w.r.t. R

Termination disproved by the backward process
proof stopped at iteration i=5, depth k=7
64 rule(s) generated