A tricky logical problem about span and linearly independency

I encounter a tricky logical problem in linear algebra. Every inference seems valid, but the result is absurd. I'm sure it will turn out that I make a mistake somewhere. However, I tried to think it for long time today(my brain's kinda short.), but can't see why.

Let $V$ be a vector space over $F$, and let $S\subseteq V$, $v\not\in S$. Then we know that

$v\not\in\text{span}(S)\Longleftrightarrow \bigl(S\text{ is linearly independent}\Rightarrow S\cup\{v\}\text{ is linearly independent}\bigr)$.

This is a biconditional. And it is of course logically equivalent to $v\in\text{span}(S)\Longleftrightarrow \neg\bigl(S\text{ is linearly independent}\Rightarrow S\cup\{v\}\text{ is linearly independent}\bigr)$.

Thus, when $v\in\text{span}(S)$, then $\neg\bigl(S\text{ is linearly independent}\Rightarrow S\cup\{v\}\text{ is linearly independent}\bigr)$, then it is again equivalent to $\bigl(S\text{ is linearly independent}\wedge S\cup\{v\}\text{ is linearly dependent.}\bigr)$ Hence we get that $S\text{ is linearly independent}$. But this is non-sense. How can $v\in\text{span}(S)$ implies $S\text{ is linearly independent}$? The consequence is too strong. Where did I make a mistake?

• You have been mistaken in understanding the statement you got. It is absolutely correct! The final result says "If the vector v belongs to the set S (which is subset of vector space V), then S is linearly independent AND S union v is linearly dependent (and vice - versa) ". This is possible because since v is a member of span of S, it can be expressed as a linear combination of vectors of S. Thus, the set S is linearly independent and S union v is linearly dependent. – Aniruddha Deshmukh Aug 29 '17 at 12:15
• Already the first logical statement is false. If $S$ is not linearly independent, then the statement in brackets is true for every choice of $v$, whereas the statement on the LHS is not always true. – Jan Bohr Aug 29 '17 at 12:15

The problem seems to be with the statement

$v\not\in\text{span}(S)\Longleftrightarrow \bigl(S\text{ is linearly independent}\Rightarrow S\cup\{v\}\text{ is linearly independent}\bigr)$

and in particular

$$v\not\in\text{span}(S)\Longleftarrow \bigl(S\text{ is linearly independent}\Rightarrow S\cup\{v\}\text{ is linearly independent}\bigr)$$ which is same as $$v\in\text{span}(S)\Longrightarrow \bigl(S\text{ is linearly independent and } S\cup\{v\}\text{ is linearly dependent}\bigr)$$ The linear dependence or independence of $S$ cannot be determined from the statement $v\in\text{Span}(S)$.

The correct statement should have been

$\bigl(v\not\in\text{span}(S) \text{ and$S$is linearly independent} \bigr)$$\Longleftrightarrow \bigl(S\cup\{v\}\text{ is linearly independent}\bigr)$

and these two statements are not equivalent.

• Thank you and @jabo. I see. :) – Eric Aug 31 '17 at 14:15