How is "~A. Therefore A -> B" a valid argument?

Question

I’m teaching myself logic from ‘Logic: The Laws of Logic’. In this book it states that

1) any sequence of propositions counts as an argument. 2) valid is when true premises of an argument cannot lead to a false conclusion.

Now look at this argument: ~A Therefore A -> B

I struggle to intuitively understand how this is a valid argument. Simply stating “not-A informs us of A’s truth value, therefore aids in understanding the conditional” does not help. Since I don’t intuitively see that not-A regulates anything about B.

E.g. Anne isn’t cool. Therefore if Anne is cool then Beth is cool.

Please explain.

Also please explain how this argument would be criticised. E.g. because it’s incomplete or??

Thank you for your time!

Answer 1

It seems like you understand how the use the formal machinery to show that ~A entails A -> B, but you're having trouble understanding what's going on (building "intuition").

Here's another approach. Remember that, in propositional logic, A -> B does not mean anything like "A causes B." So you can't rely on your understanding of causal relationships to get an understanding of how A -> B works. Instead, by definition, A -> B means exactly ~A v B, "either not-A or B." Nothing more or less. (This is called the "material conditional." The Stanford Encyclopedia has an entry on the logic of conditionals that's demanding for a beginner but might be very interesting for you.)

Whenever you have A -> B, you can replace it with ~A v B. And vice versa. So, in your argument, the question is whether ~A entails ~A v B. If A is false, can we be certain that either A is false or B is true? Hopefully it's clear that the answer is "yes."

Part of the trick to mastering formal logic is recognizing when the formalism works like natural language ("or" is pretty similar) and when it works completely differently (the material conditional). The formalism is extremely powerful, but that power comes from satisfying some very narrow requirements that can have confusing implications.

Answer 2

If A is assumed to be False (this is the meaning of the premise ~A) we can correctly conclude that A → B is True.

Consider the truth table for the conditional, we have that it is True when the antecedent is False, irrespective of the truth value of the consequent, i.e. :

when A is False, A → B is True, whatever B is.

See the definition of Logical consequence :

"every tuth assignment that causes all the premises to be true also assign true to the conclusion".

But we have only one choice for the truth assignment that cause the premise to be True: it must assig to A the value False.

This choice forces us to select the two last rows of the truth table, and we have that in these rows the conclusion is marked True.

Thus, the argument is valid, because A → B is a logical consequence of ~A.

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See also Ex falso sequitur quodlibet.

Answer 3

I agree with Mauro ALLEGRANZA's answer. I am providing this answer to complement it.

Since you are learning logic on your own I will illustrate a solution to this problem using forall x: Calgary Remix and Kevin Klement's natural deduction proof checker. These are available online and could supplement the text you are using. They cover truth-functional logic and first order logic.

You have a truth-table as one way to show the validity of the argument. Here is a derivation illustrated with a natural deduction Fitch-style proof checker:

On line 1, the premise "¬A" is stated. This was automatically provided once I entered the premise and the conclusion in the proof checker.

On line 2, I assume "A". I can assume anything. Ideally I want to assume something useful in achieving the goal which is "A→B". This assumption is indented because the proof checker uses the "Fitch-style".

On line 3, I note that line 1 and line 2 are contradictory, so I introduce (I) a contradiction (⊥) and reference lines 1 and 2 (⊥I 1, 2). The software will make sure I get this correct. If it is not when I click "Check Proof" it will show an error and not let me reach the "Congratulations!" confirmation message.

On line 4, I use something that should seem shocking. It is called "explosion" (X). If I have a contradiction, I can then place on this line anything. Since I need "B", I put "B" there and justify it using explosion and reference line 3 where I introduced the contradiction (X 3).

On line 5, I "discharge" the assumption I made on line 2 by restating that indented subproof as a conditional, if "A" then "B" or "A→B", and justify that line by using (→I 2–4) where I introduce (I) the conditional (→) referring to a string of lines starting at 2 and ending at 4. This new line can be thought of as a summary of those three lines.

Beyond the truth table and the natural deduction argument, you might still ask why should this work?

Think of "A" as the symbolization for this English sentence: "The moon is made of green cheese." Clearly, that is not the case. So "¬A" is true. Now assume that the moon is made of green cheese. To make this even more concrete, suppose I'm writing a novel with a green-cheese moon. Then what? Then anything can happen that I want to happen.

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References

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

Wikipedia, "Fitch notation" https://en.wikipedia.org/wiki/Fitch_notation

Answer 4

To understand a proof of ~A => [A => B], you will have to be familiar with the methods of direct proof, proof by contradiction and removal of double negation. (See my formal proof here.) Otherwise, you will have to rely on the "definition" of '=>' given by the last 2 lines of the truth table for A => B shown here. Note that if A is false, then A=>B is true regardless of whether B is true or false.