# Is this proof valid

This is a clear version of an earlier closed question.

Let: $x, u, y, v : \mathbb{C} \to \mathbb{R}$ be functions in the complex variable $s=a+ib$ defined by convergente series. I have this implication:

If $x(s)=u(s), y(s)=v(s)$, then after some steps on these series one get an equation of the form: $h(s)=0$ which give one value of $s$ as $s=1+ib$ for some $b$.

I call this result ($s=1+ib$) as "(P)".

The function $h$ here depends on the given functions $x, u, y, v$. Its formula is not important.

let $A={s∈\mathbb{C}: x(s)=u(s),y(s)=v(s)}$ and $B= {s∈\mathbb{C}: x(s)=u(s)=y(s)=v(s)=0}$ is a proper subset of $A$.

I know that if $s∈A$ then (P) holds true.

My question is as follow: What happen if $s∈B$. Is (P) true or false!.

We remark here that if $s∈B$ then we cannot construct the function $h$ and hence we cannot deduce (P), i.e., if $x(s)=u(s)=y(s)=v(s)=0$ then I cannot do anything and the proof stoped here.

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If $s\in B$, then $s\in A$, as $B$ is a subset of $A$. So (P) must be true. However, if you cannot construct the function $h$ for $s\in B$, then (P) makes no sense in this case and can't be true, and therefore it doesn't follow from $s\in A$ either. – Matthew Pressland Jan 8 '13 at 18:50
French valable is English valid. – Brian M. Scott Jan 8 '13 at 18:51
I can't tell without knowing the formula for $h$, but my guess would be you need to additionally require that some function of $x,u,y,v$ is non-zero in the definition of $A$, thus excluding some points of $B$. – Matthew Pressland Jan 8 '13 at 18:52
if $x(s)=u(s)=y(s)=v(s)=0$ then I cannot do anything and the proof stoped here. – ZE1 Jan 8 '13 at 18:55
@RH1 But then $x(s)=u(s)$ and $y(s)=v(s)$, so the rest of the proof should be exactly the same as what you claim to have already proved. If this doesn't work, there is a problem with the original proof of (P). – Matthew Pressland Jan 8 '13 at 18:59

As Matt Pressland hinted: Since we have still $x(s)=u(s)$ and $y(s)=v(s)$, then the rest of the proof is exactly the same as in the case of the set $A$. So, the implication is true and my idea saying that if $x(s)=u(s)=y(s)=v(s)=0$ then I cannot do anything is wrong. The principle here is the equations $x(s)=u(s)$ and $y(s)=v(s)$ not their exact values.