# Rule of inference in logic

There is a slide in my class notes that mention

$$\neg p \rightarrow F_0$$

therefore p.

It then follows up and it says

if we want to establish the validity of the argument

Eqn 1: ($$p_0 \wedge p_1... \wedge p_n)$$ -> q,

we can establish the validity of logically equivalent argument.

Eqn 2: ($$p_0 \wedge p_1... \wedge p_n \wedge \neg q) -> F_0$$

My questions:

1. The implication

$$\neg p \rightarrow F_0$$

leads to "therefore p" because p is assumed to be true and the only way p can lead to false is through negation of p? I'm very confused about this

1. How do we go from the first equation to the second?
• Maybe contraposition? – jMdA Feb 2 at 5:15

Assuming that $$F_0$$ means $$\bot$$ (the always False logical constant), we can easily check (with truth table) that $$p \to F_0$$ is equivalent to $$\lnot p$$: due to the fact that $$F_0$$ is always False, we have only two cases: $$\text T → \text F$$ and $$\text F → \text F$$, and the result is exactly the opposite of $$p$$.
Thus, $$\lnot p \to F_0$$ is $$\lnot \lnot p$$ that, in classical logic, is equivalent to $$p$$.
For the same reason, $$(p_i \land \lnot q) \to F_0$$ is equivalent to $$\lnot (p_i \land \lnot q)$$ which is in turn equivalent to $$p_i \to \lnot \lnot q$$.
• @user162264 - NO; $F_0$ is ALWAYS False. Thus we have only two cases $T \to F$ and $F \to F$. Compute them... – Mauro ALLEGRANZA Feb 2 at 7:08