We're covering Prepositional Logic in our class and I'm a little confused as to how the entire concept actually works. My understanding is that prepositional logic is used to study ways of combining or altering statements or propositions to form more complicated statements or propositions. Therefore when we have logical expression, we have to show that the statement is true or false.
Let's say I want to show that the following logical statement is true:
$$(\exists x. P_1(x) \to P_2(x)) \to ((\forall y.P_1(y)) \to \exists z.P_2(z))$$
- Assumption ($A1$): $(\exists x. P_1(x) \to P_2(x))$
- To Show ($S1$): $((\forall y.P_1(y)) \to \exists z.P_2(z))$
- Assumption ($A2$): $((\forall y.P_1(y))$
- To Show ($S2$): $\exists z.P_2(z))$
Now I know that at this point I should choose $x$ as arbitrary but fixed in A1, but I don't know the exact reason for this, instead of lets says choosing $z \equiv x$.
- Assumption ($A3$): $P_1(x) \to P_2(x)$
- Choose $y \equiv x$ in A2.
- Assumption ($A4$): $P_1(x)$
After this point, I'm uncertain with what to do.
PS - I know that "To Show" and "Assumption" may not be the correct English terms but they're the closest translation from German "Annahme" and "Zu Zeigen" respectively.
Here is an outline of an argument I assume you can turn into whatever rigor you need:
1) If $\exists x. P_1(x)\rightarrow P_2(x)$ is false, we are done. If it is true, then there is a witness to that statement being true, i.e. there is a $c$ such that $P_1(c)\rightarrow P_2(c)$.
2) If $\forall y. P_1(y)$ is false, then we are done. If it is true, then in particular $P_1(c)$ is true.
3) Since we are in the case that $P_1(c)\rightarrow P_2(c)$ is true and that $P_1(c)$ is true, by the comment made by Git Gud we know that $P_2(c)$ is true by modus ponens.
4) Since $P_2(c)$ is true, it must be the case that $\exists z. P_2(z)$ is also true.
5) Now everything in sight is true, and the cases where anything ended up being false have been addressed, so the statement is logically true.