Does this implication involving real numbers $a$, $b$ hold?

Question

Suppose $a$ and $b$ are real numbers that satisfy all of the following properties:

1. $a\neq 0$
2. It is not the case that both $a<0$ and $b<0$. Symbolically, $\neg(a<0\wedge b<0)$.
3. It is not the case that both $a=0$ and $b=0$. Symbolically, $\neg(a=0\wedge b=0)$.
4. $b\leq 0$ implies $a>0$. Symbolically, $b\leq 0\Rightarrow a>0$.

Does it necessarily follow that either ($a>0\wedge b\in\mathbb{R}$) or ($a<0\wedge b>0$), but not both?

I tried reasoning through this by making a list of all the possibilities for $a$ and $b$.

$$a<0\wedge b<0$$ $$a<0\wedge b=0$$ $$a<0\wedge b>0$$ $$a=0\wedge b<0$$ $$a=0\wedge b=0$$ $$a=0\wedge b>0$$ $$a>0\wedge b<0$$ $$a>0\wedge b=0$$ $$a>0\wedge b>0$$

My first insight was that $b\leq 0\Rightarrow a>0$ is logically equivalent to $a\leq 0 \Rightarrow b>0$, so lines 1, 2, 4 and 5 cannot possibly be true. This leaves

$$a<0\wedge b>0$$ $$a=0\wedge b>0$$ $$a>0\wedge b<0$$ $$a>0\wedge b=0$$ $$a>0\wedge b>0$$

I then realized that $a\neq 0$ makes $a=0$ false, so the conjunction $a=0\wedge b>0$ is necessarily false. This leaves me with four possibilities:

$$a<0\wedge b>0$$ $$a>0\wedge b<0$$ $$a>0\wedge b=0$$ $$a>0\wedge b>0$$

At this point, I'm stuck. I'm not sure if the four properties listed above necessitate the truth of exactly one of these four conjunctions, and if they do, how to prove it. Could I get some assistance?

Update: I found a way to justify this implication! You see, for any pair of numbers $\{a,b\}$, it must be the case that exactly one of the nine original conjunctions is true. By eliminating the 5 that I did, I inadvertently showed that exactly one of the four remaining ones must be true, which is what I sought to prove. I'm no longer seeking answers, but if you have something you feel will be beneficial to me, please feel free to respond.

Answer 1

Your logic is correct. Now you end up with

$$(a<0\land b>0) \lor (a>0\land b<0)\lor (a>0\land b=0)\lor (a>0\land b>0)$$

Using distributivity rules, the above is equivalent to

$$(a<0\land b>0) \lor (a>0\land (b<0\lor b=0\lor b>0))$$

Use the law of trichotomy, the condition $b<0\lor b=0\lor b>0$ is equivalent to $b\in\mathbb R$.

This leads to

$$(a<0\land b>0) \lor (a>0\land b\in\mathbb R)$$

To show exclusive or, we consider the conjuction

$$(a<0\land b>0) \land (a>0\land b\in\mathbb R) \equiv a<0\land a>0 \land b>0\land b\in\mathbb R$$

and we see that $a<0 \land a>0$ is false. Hence the OR is exclusive.