Implication statements involving a variable

Solution 1:

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 $$\exists x\,\big(x = 1 \;\text{and}\; x \ne 2\big)\tag{1n}$$ $$\exists x\,\big(x\geq0 \;\text{and}\; \textrm{abs}(x) \leq0\big)\tag{2n}.$$

Solution 2:

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.