$\exists x\,\big( x >0 → x^3 < 0\big)$ I am very confused on learning mathematical logic.
The problem is to determine whether the statement is true or false.
Statement: $\exists x ( x >0 \rightarrow x^3 < 0)$ is true. (the domain is $\mathbb R$.)
I think it is false since there is no $x >0 $ such that $x^3 <0$. However, I am confused on the other solution.
The other solution is:
If predicates $P(x): x >0$ and $Q(x): x^3 <0$ are defined, then $P(-1)$ is always false.
Thus, the predicate $P(-1) \rightarrow Q(-1)$ is also true which implies that there is $x$ such that $x >0 \rightarrow x^3 < 0$ is true. As a result, the statement is true.
I also know that the second solution is false, but I do know know how to exactly describe the faults.
 A: Let the universe of discourse be $\mathbb R.$

*

*$$\forall x\;\big( x >0 \land x^3 < 0\big)$$    Every $x$ is positive and has a negative cube.
False.


*$$\exists x\;\big( x >0 \land x^3 < 0\big)$$    Some positive $x$ has a negative cube.
More literally: some $x$ is positive and has a negative cube.
False.


*$$\forall x\;\big( x >0 \rightarrow x^3 < 0\big)$$    Every positive $x$ has a negative cube.
More literally: every $x$ that is positive has a negative cube.
False (disprove by finding a true antecedent and false consequent).


*$$\exists x\;\big( x >0 \rightarrow x^3 < 0\big)$$    For some $x,$ if it's positive, then its cube is negative.
This translation is the most literal among the four; it is also the trickiest,  as we are tempted to conflate the symbolic versions of $(2)$ and $(4);$ fortunately, this sentence category is also the least common.
True, as explained by your 2nd solution (prove by finding a false antecedent or true consequent). Clarify the symbolic sentence by equivalently rewriting it as $$\exists x\;\big( x \leq0 \,\lor\, x^3 < 0\big).$$
