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?
  • $\begingroup$ Maybe contraposition? $\endgroup$ – 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 $p_i \to \lnot \lnot q$.

  • $\begingroup$ Just so I understand the first point: p-> F_0 means when p is true, F_0 is false and when p is False, F_0 is true. Thus, since F_0 is the opposite of p, this overall statement (p->F_0) is equivalent to -p ? $\endgroup$ – user162264 Feb 2 at 7:06
  • 2
    $\begingroup$ @user162264 - NO; $F_0$ is ALWAYS False. Thus we have only two cases $T \to F$ and $F \to F$. Compute them... $\endgroup$ – Mauro ALLEGRANZA Feb 2 at 7:08

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