Is there an explanation that would make this proof more understandable?

Question

From an answer to another question:

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$¬A⟹[A⟹B]$

Proof

1. Suppose $¬A$

2. Suppose $A$

3. Suppose (to the contrary) $¬B$

4. Obtain the contradiction $A∧¬A$ from (2) and (1).

5. Conclude by contradiction that $¬¬B$ from (3) and (4).

6. Remove $¬¬$ to obtain $B$ from (5).

7. Conclude that $A⟹B$ from (2) and (6).

8. Conclude as required that $¬A⟹[A⟹B]$ from (1) and (7).

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I have spent the last month reading about sentence logic. I am looking for proof (for lack of a better word) that if $P$ and $Q$ are both false, $P \implies Q$ is true. The above proof supposedly is what I am looking for and I am not arguing that it is not. I am looking for a more annotated proof.

For example, line 1 states: Suppose $¬A$. This seems straight forward, as do lines $2$ and $3$. Line $4$ is understandable. It is as if I said go left and go right at the same time. You can not do both. Likewise, $A$ can not be both true and false at the same time. Of the remaining lines, only line $6$ is clear.

This is, of course, what I am not seeing or understanding. I have read a lot, searched through questions on this site and filled half a small notebook with scribbling and questions. I can work through the exercises in most of the texts I have read. None of them are very complex. None of them have required a proof other than using a truth table as a demonstration.

Am I unable to follow or understand the proof above because I need to read additional chapters or because there is something advanced in the text of the proof or how that text is presented? Sorry to clutter up the list of questions with this. I do not need a detailed explanation of the proof as an answer so much as a reference to where I would find an answer.

Answer 1

Long comment

The issue it seems to be that the proof (wherever you found it) is not correctly annotated: every step must be commented with the name of the rule used with ref to the previous lines to which the rule has been applied.

Using the Natural Deduction proof system (the proof is "by contradiction": assume the negation of the sought conclusion and derive a contradiction):

1) $¬A$ --- assumption

2) $A$ - assumption

3) $¬B$ --- assumption

4) $A \land ¬A$ --- from (2) and (1) by Conjunction Introduction

5) $¬¬B$ --- from (3) and (4) by Negation Introduction, discharging assumption (3)

6) $B$ --- from (5) by Double Negation

7) $A \to B$ --- from (2) and (6) by Implication Introduction (aka: Conditional Proof), discharging assumption (2)

8) $¬A \to (A \to B)$ --- from (1) and (7), concluding as required by Implication Introduction again, discharging assumption (1).

Answer 2

Suppose to the contrary, that

$$\lnot (((\lnot P)\land(\lnot Q))\to(P\to Q)).\tag{1}$$

Then $(\lnot P)\land(\lnot Q)$ is true while $(P\to Q)$ is false; the former gives $\lnot P$ (and $\lnot Q$), whereas the latter gives $P$ (and $\lnot Q$). This is a contradiction.

The negation of $(1)$ must, then, follow.$\square$

Visually:

This is a proof tree.