Which case (1) or (2) that disprove case $(A)$?

Question

I have the following logical case:

(A): $$ \exists n \geq 1, \forall m \geq n, \forall j \geq n, \forall a, b, c,d,e,z,r,t,p \in \mathbb{N}: f(m,j,a,b,c,d,e,z,r,t,p)>0$$

where $f$ is a positive function.

I want to show that this case if false. Then I proceed as follow:

(1) I find $m$ and $j$ depending on $n$ and some fixed $a,b,c,d,e,z,r,t,p$ such that $$ f(m(n),j(n),a,b,c,d,e,z,r,t,p)=0$$ for all $n$.

(2) I find $m$ and $j$ depending on $n$ and some fixed $a,b,c,d,e,z,r,t,p$ such that $$ f(m(n),j(n),a,b,c,d,e,z,r,t,p)=0$$ for infinitely many $n$.

Hence, my question is: Which case (1) or (2) that disprove case $(A)$.

Answer 1

Obviously the first one: the negation of "there is some n such that P" will be "for every n, not-P".

Assume that for even values of $n$ we can "find $m$ and $j$ depending on $n$ and some fixed $a$" such that:

$f(m(n),j(n),a)=0$, for all [even] $n$.

Thus, $f(m(n),j(n),a)=0$ will hold for infinitely many $n$.

But assume that for, e.g. $n=3$ we cannot find suitable $m,j,a$.

This fact is consistent with $∃n≥1,∀m≥n,∀j≥n,∀a: f(m,j,a)>0$