Does a statement p that's always false imply that p implies any statement q?

Question

Question

Is the following statement true?

For all statements p, p always being false implies that for all statements q, p implies q, i.e:

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

If so, why? If not, why not?

Thanks.

Attempt

Here's my attempt of explaining why I think that the above statement is true. I am new to this notation, so pardon me if I made any mistakes.

Assumption 1:

For all statements p, p does not imply q when there exists a statement r, where r implies p but also implies not q.

$$\forall p(p \nRightarrow q) \equiv \exists r((r \implies p) \land (r \implies \lnot q))$$

Assumption 2:

If we've disproven that p does not imply q, then p implies q.

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

Proof:

$p$ is always false, therefore there does not exist $r$ such that $r \implies p$

Therefore, there does not exist $r$ such that $(r \implies p) \land (r \implies \lnot q)$

Therefore, based on assumption 1 above, the statement $p \nRightarrow q$ is false

Therefore, based on assumption 2 above, $p \implies q$

However, my two assumptions could be wrong! It's also messy that I am defining logic using logic.

Context

Here is a proof I was told for the following statement in relation to sets, which goes on the assumption that my statement at the beginning of this post is true.

Statement to prove:

$$\forall A(\emptyset \subset A)$$

A definition for sets:

$$A \subset B \equiv \forall a(a \in A \implies a \in B)$$

Proof:

$$\emptyset \subset A \equiv \forall x(x \in \emptyset \implies x \in A)$$

$x \in \emptyset$ is never true, therefore, based on the statement at the beginning of this post, $x \in \emptyset \implies b \in \emptyset$ is always true, therefore $\emptyset \subset A$

Your translations aren't quite correct; e.g., the first one ought to be $\forall p\bigg( \lnot p \implies \big(\forall q( p \implies q)\big)\bigg)$ instead. Are you using $\equiv$ and $\Leftrightarrow$ synonymously? Related: False implies anything — ryang, Sep 28 '21 at 03:57
Yes, $\lnot p \to (p \to q)$ is a tautology (check it with truth table). — Mauro ALLEGRANZA, Sep 28 '21 at 07:12
But it is not correct to say that $\lnot p$ represent a statement that is always false. A contradiction, like $p \land \lnot p$ is always false. In prop logic we can symbolize with $\bot$ a statement that is always false. — Mauro ALLEGRANZA, Sep 28 '21 at 07:13
Regarding $\emptyset \subseteq A$, for every $A$, your argument is correct. — Mauro ALLEGRANZA, Sep 28 '21 at 07:14
@RyanG Yes I am using $\equiv$ and $\Leftrightarrow$ synonymously. See "material equivalence" in this section on Wikipedia. — David Callanan, Sep 28 '21 at 11:23

Dan Christensen · Accepted Answer · 2021-09-28T16:50:03.660

1

For any propositions $P$ and $Q$, here is the truth table for $P\to (\neg P \to Q)$

Source: https://www.erpelstolz.at/gateway/TruthTable.html

Here a proof using a form of natural deduction. For any propositions $P$ and $Q$, we have:

Source: Screenshot from my proof checker

edited Sep 28 '21 at 16:50

answered Sep 28 '21 at 16:31

Dan Christensen

14,529

Does a statement p that's always false imply that p implies any statement q?

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