Implication statements involving a variable I just learned the definition of implication and have been considering a few problems.
Given $x$ is real, prove $x = 1 \Rightarrow x = 2$.
Either $x = 1$ or $x \neq 1$. If $x=1$, then the implication is false, and if $x \neq 1$ is true, then the implication is true. Since we don't know the value of $x$, then this statement cannot be proved.
Also consider given $x$ is real, prove it is false that $x \ge 0 \implies \text{abs}(x) > 0$.
Textbook answers usually go: when x is $0$, $\text{abs}(x)$ is $0$ and therefore the statement is wrong.
But isn't this the proof of the NEGATION of the UNIVERSAL statement (for all real $x$, if $x\le 0$, then $\text{abs}(x) > 0$) by proving there exists a value of $x$ (i.e. 0) such that $\text{abs}(x)$ is not greater than $0$?
To prove this statement false, we need to prove $x \ge 0$ AND $\text{abs}(x) \le 0$. But we don't know the value of $x$, so rigorously speaking, we cannot prove this, right?
Maybe there is some sort of convention regarding this question?
However,
suppose now the question is instead if $x > 0$, prove $\text{abs}(x) >0$. In this scenario, to me, it is clear that it is the question about this SPECIFIC value of x that we do not know that is being asked (it just sounds right, and you often see problems start with an unknow e.g. the possibility of something is p...), even though the universal statement about all $x$ implies this and is probably most people would use to deduce this statement.
So it seems like there is some convention regarding proving the falsehood of such a statement, but not for the truth?
 A: Hint 1: Proving $\neg \forall x:[x\in R \to [x=1 \to x=2]]$
Suppose to the contrary. Assuming $~1\in R~$ and $~1\neq 2$, obtain a contradiction.

Hint 2: Proving $\neg \forall x:[x\in R \to [x\leq 0 \to |x|\gt0]]$
Suppose to the contrary. Assuming $~0\in R, ~|0|=0,~0\leq 0$ and $~0 \ngtr  0$, obtain a contradiction.
A: 
Given that $x$ is real, prove that $$x = 1 \implies x = 2.\tag1$$
Since we don't know the value of x, then this statement cannot be proved.


given that $x$ is real, prove that it is false that $$x\geq0\implies\textrm{abs}(x) >0.\tag2$$
To prove that this statement false, we need to prove that $x\geq0$ AND $\textrm{abs}(x) \leq 0.$ But we don't know the value of $x,$ so, rigorously speaking, we cannot prove this, right?

$(1)$ and $(2)$ are open formulae and thus not statements, let alone provable statements. However, if we treat them as being implicitly universally quantified, i.e., $$\forall x\,\big(x = 1 \implies x = 2\big)\tag{1a}$$ and $$\forall x\,\big(x\geq0\implies\textrm{abs}(x) >0\big)\tag{2a},$$ then the counterexamples $x=1$ and $x=0,$ respectively, show that they are false statements.

If asked to prove something is false, should I prove that statement is false for all values of variables (x, y, z, etc.), or should I prove the falsehood of the universal statement involving all the variables(i.e. to prove there exists some values that the statement is false)?

The latter. Providing the counterexamples above is essentially disproving $(1\mathrm a)$ and $(2\mathrm a)$ by proving their negations $$\exists x\,\big(x = 1 \;\text{and}\; x \ne 2\big)\tag{1n}$$ $$\exists x\,\big(x\geq0 \;\text{and}\; \textrm{abs}(x) \leq0\big)\tag{2n}.$$
