Two ways to represent ex falso?

Question

According to ProofWiki, the principle of explosion says that:

$$p \implies (\lnot p \implies q)$$

but sometimes it's also stated as

$$(p \land \lnot p) \implies q$$

Can one be converted to the other? How do you prove or show these?

If we know that $(p \implies q) \equiv \lnot p \lor q$ by definition, then substituting the 2nd representation:

$$((p \land \lnot p) \implies q) \equiv \lnot(p \land \lnot p) \lor q \equiv \lnot p \lor p \lor q$$

Is this just true due to law of excluded middle? Like either $p$ or $\lnot p$ is true, i.e. $p \lor \lnot p \equiv \text{T}$ before we even get to whatever $q$ is. Is that why we're able to say it?

Then the first representation is weird too:

$$(p \implies (\lnot p \implies q)) \equiv (\lnot p \lor (\lnot p \implies q)) \equiv (\lnot p \lor (p \lor q)) \equiv \lnot p \lor p \lor q$$

I mean am I even on the right track here? How do you normally get these without "working backwards" and showing that they happen to be equivalent? Is there a more intuitive derivation of both of these?

Answer 1

The formulas $p \to (\lnot p \to q)$ and $(p \land \lnot p) \to q$ are equivalent because they are particular instances of a more general fact: the formulas $p \to (r \to q)$ and $(p \land r) \to q$ are equivalent. You can prove that checking that the two formulas actually have the same truth table. An alternative proof is the following:

\begin{align} p \to (r \to q) &\iff \lnot p \lor (\lnot r \lor q) &\text{by definition of } \to \\ &\iff (\lnot p \lor \lnot r) \lor q &\text{by associativity of } \lor \\ &\iff \lnot(p \land r) \lor q &\text{by De Morgan law} \\ &\iff (p \land r) \to q &\text{by definition of }\to \end{align}

Answer 2

The two statements you post are logically equivalent.

$$\begin{align} p \to (\lnot p \to q) &\equiv \lnot p \lor (p \lor q) \tag{definion of $\to$}\\ \\ &\equiv (\lnot p \lor p) \lor q \tag{associativity of $\lor$}\\ \\ &\equiv \lnot (p \land \lnot p) \lor q \tag{DeMorgan's}\\ \\ &\equiv (p\land \lnot p)\to q\tag{definition of $\to$}\end{align}$$

You were heading in right direction, starting instead from $(p\land \lnot p)\to q$ and moving toward $p \to (\lnot p \to q)$. You just stopped too early. Note each direction can be proven, because we have a string of equivalencies: so to reverse the direction, just reverse the order of the steps.

$$\begin{align} (p\land \lnot p)\to q &\equiv \lnot (p \land \lnot p)\lor q\tag{definition of $\to$}\\ \\ &\equiv (\lnot p \lor p) \lor q\tag{DeMorgan's}\\ \\ &\equiv \lnot p\lor (p \lor q)\tag{associativity of $\lor$}\\ \\ &\equiv p\to (\lnot p \to q)\tag{definition of $\to$} \end{align}$$

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First of all, we have $$p\to (q\to r) \vdash (p \land q) \to r\tag{importation}$$ and we have $$(p\land q) \to r \vdash p \to (q\to r)\tag{exportation}$$ Your question is a specific case of these, where $q = \lnot p$. See this link for more information.

Answer 3

To get pernickety, we need to distinguish the principle about entailment what is usually called Explosion, i.e.

$(A \land \neg A) \vdash B$,

from the following principle that a conditional with a contradictory antecedent is always derivable:

$\vdash (A \land \neg A) \to B$.

You can have logics with one principle and not the other.