let R be the TRS under consideration

f(g(_1),_2) -> f(_1,n__f(n__g(_1),activate(_2))) is in elim_R(R)
let r0 be the right-hand side of this rule
p0 = epsilon is a position in r0
we have r0|p0 = f(_1,n__f(n__g(_1),activate(_2)))
f(_3,_4) -> n__f(_3,_4) is in R
let l'0 be the left-hand side of this rule
theta0 = {_1/_3, _4/n__f(n__g(_3),activate(_2))} is a mgu of r0|p0 and l'0

==> f(g(_1),_2) -> activate(_2) is in EU_R^1
let r1 be the right-hand side of this rule
p1 = epsilon is a position in r1
we have r1|p1 = activate(_2)
activate(n__f(_3,_4)) -> f(activate(_3),_4) is in R
let l'1 be the left-hand side of this rule
theta1 = {_2/n__f(_3,_4)} is a mgu of r1|p1 and l'1

==> f(g(_1),n__f(_2,_3)) -> f(activate(_2),_3) is in EU_R^2
let r2 be the right-hand side of this rule
p2 = 0 is a position in r2
we have r2|p2 = activate(_2)
activate(n__g(_4)) -> g(activate(_4)) is in R
let l'2 be the left-hand side of this rule
theta2 = {_2/n__g(_4)} is a mgu of r2|p2 and l'2

==> f(g(_1),n__f(n__g(_2),_3)) -> f(g(activate(_2)),_3) is in EU_R^3
let r3 be the right-hand side of this rule
p3 = 0.0 is a position in r3
we have r3|p3 = activate(_2)
activate(_4) -> _4 is in R
let l'3 be the left-hand side of this rule
theta3 = {_2/_4} is a mgu of r3|p3 and l'3

==> f(g(_1),n__f(n__g(_2),_3)) -> f(g(_2),_3) is in EU_R^4
let r4 be the right-hand side of this rule
p4 = epsilon is a position in r4
we have r4|p4 = f(g(_2),_3)
f(g(_4),_5) -> f(_4,n__f(n__g(_4),activate(_5))) is in R
let l'4 be the left-hand side of this rule
theta4 = {_2/_4, _3/_5} is a mgu of r4|p4 and l'4

==> f(g(_1),n__f(n__g(_2),_3)) -> f(_2,n__f(n__g(_2),activate(_3))) is in EU_R^5
let r5 be the right-hand side of this rule
p5 = 0 is a position in r5
we have r5|p5 = _2
activate(n__g(_4)) -> g(activate(_4)) is in R
let l'5 be the left-hand side of this rule
theta5 = {_2/activate(n__g(_4))} is a mgu of r5|p5 and l'5

==> f(g(_1),n__f(n__g(activate(n__g(_2))),_3)) -> f(g(activate(_2)),n__f(n__g(activate(n__g(_2))),activate(_3))) is in EU_R^6
let l be the left-hand side and r be the right-hand side of this rule
let p = epsilon
let theta = {}
let theta' = {_1/activate(_2), _2/_2, _3/activate(_3)}
we have r|p = f(g(activate(_2)),n__f(n__g(activate(n__g(_2))),activate(_3))) and
theta'(theta(l)) = theta(r|p)
so, theta(l) = f(g(_1),n__f(n__g(activate(n__g(_2))),_3)) is non-terminating w.r.t. R

Termination disproved by the forward process
proof stopped at iteration i=6, depth k=3
366 rule(s) generated