let R be the TRS under consideration

incr(cons(_1,_2)) -> 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 = activate(_2)
activate(n__incr(_3)) -> incr(activate(_3)) is in R
let l'0 be the left-hand side of this rule
theta0 = {_2/n__incr(_3)} is a mgu of r0|p0 and l'0

==> incr(cons(_1,n__incr(_2))) -> incr(activate(_2)) is in EU_R^1
let r1 be the right-hand side of this rule
p1 = 0 is a position in r1
we have r1|p1 = activate(_2)
activate(n__oddNs) -> oddNs is in R
let l'1 be the left-hand side of this rule
theta1 = {_2/n__oddNs} is a mgu of r1|p1 and l'1

==> incr(cons(_1,n__incr(n__oddNs))) -> incr(oddNs) 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 = oddNs
oddNs -> incr(pairNs) is in R
let l'2 be the left-hand side of this rule
theta2 = {} is a mgu of r2|p2 and l'2

==> incr(cons(_1,n__incr(n__oddNs))) -> incr(incr(pairNs)) 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 = pairNs
pairNs -> cons(0,n__incr(n__oddNs)) is in R
let l'3 be the left-hand side of this rule
theta3 = {} is a mgu of r3|p3 and l'3

==> incr(cons(_1,n__incr(n__oddNs))) -> incr(incr(cons(0,n__incr(n__oddNs)))) is in EU_R^4
let l be the left-hand side and r be the right-hand side of this rule
let p = 0
let theta = {_1/0}
let theta' = {}
we have r|p = incr(cons(0,n__incr(n__oddNs))) and
theta'(theta(l)) = theta(r|p)
so, theta(l) = incr(cons(0,n__incr(n__oddNs))) is non-terminating w.r.t. R

Termination disproved by the forward process
proof stopped at iteration i=4, depth k=3
1727 rule(s) generated