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So I am asked to put together a tautology that accurately reflects a substitution instance in First Order Logic for the the following sentence:

“If I have two dollars, I can buy a soda, and if I have two dollars, then I can buy a candy bar, so if I have two dollars, I can buy a soda and a candy bar”

I have come up with the following: ((A → B) ∧ (A → C)) → (A → (B ∧ C))

Then I am asked to give an informal proof that demonstrates that it is a tautology. How do I do an informal proof?

Ram
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    "Informal proof" ? Maybe truth table ? Or a semantical argument by contradiction ? – Mauro ALLEGRANZA Dec 13 '17 at 15:38
  • Assume is not a tautology; then there is a valuationn$v$ such that: $v((A → B) ∧ (A → C))=$ TRUE and $v((A → (B ∧ C))=$ FALSE... – Mauro ALLEGRANZA Dec 13 '17 at 15:46
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    The funny thing is that of course the argument is invalid: what if both the candy bar and the soda cost exactly two dollars? Indeed, to say "I can buy a candy bar and a soda" is not the same statements as "I can buy a candy bar and I can buy a soda", so I would object to your symbolization. – Bram28 Dec 13 '17 at 15:50
  • “If I have two dollars, I can buy a soda, and if I have two dollars, then I can buy a candy bar, so if I have four dollars, I can buy a soda and a candy bar” To stipulate a lower price to buy both products requires further information since all we know currently is that the cost of the soda+candy bar is $\le$$4$ – James Arathoon Dec 13 '17 at 15:59
  • Isn't this a non sequitur rather than a tautology because the conclusion/inference could either be true or false, but the argument demands the inference definitely true nonetheless https://en.wikipedia.org/wiki/Non_sequitur_(logic) – James Arathoon Dec 13 '17 at 16:08
  • If you negate the above inference “If I have two dollars, I can buy a soda, and if I have two dollars, then I can buy a candy bar, so if I have two dollars, I cannot buy a soda and a candy bar” you do not end up with a contradiction but just another non sequitur. – James Arathoon Dec 13 '17 at 16:14
  • James, consider changing "buy" to "afford" and rephrasing the consequent of the conclusion as "I can afford a soda, and I can afford a candy bar". Then the argument can be understood as a sound application of classical logic. For it to be a non-sequitur we need to interpret it in modal logic, that is, "If I have two dollars, then it is possible for me to buy both a candy bar and a soda" does not follow from the premises. – Dan Brumleve Dec 13 '17 at 16:35
  • @Dan Brumleve: Ignoring of course the distinction between buy and afford and whether or not the shop keeper will sell both products to a single buyer. The non-sequitur includes the inference, which you have not included. Are you seriously arguing that you cannot create a non sequitur using classical logic? – James Arathoon Dec 13 '17 at 17:09
  • I'm saying that the idea of buying two things at the same time isn't really expressible in classical logic. So while ((A → B) ∧ (A → C)) → (A → (B ∧ C)) is a tautology, it doesn't capture that meaning. But it's not technically a non-sequitur, since (A → (B ∧ C)) indeed follows from ((A → B) ∧ (A → C)). – Dan Brumleve Dec 13 '17 at 17:13
  • On the other hand, if with a modal formulation we argued that $((A \rightarrow \square B) \land (A \rightarrow \square C))$ implies that $A \rightarrow \square (B \land C)$, where $\square$ denotes possibility, then that argument would be a non-sequitur. – Dan Brumleve Dec 13 '17 at 17:15
  • Let us continue this discussion in chat. – Dan Brumleve Dec 13 '17 at 17:17
  • @DanBrumleve: Sorry I had to sort out a few other things. Since [If A then choose (X OR Y)] (1) AND [If 2A then choose (X AND Y)] (2) can be applied successfully in this case in both first order logic and the logic of possibility, then it is a basic contradiction to apply [If A then choose (X AND Y)] (3) to this situation. In any system of logic (1) (2) and (3) can't all be correct. – James Arathoon Dec 13 '17 at 19:57
  • If the OP had not referred to a system of logic we could argue that the statement was provably true without contradiction in an artificially limited context and then and only then could we claim it to be a tautology (the set of tautologies includes valid mathematical deduction from an agreed set of assumptions or axioms). Why should we include this paradoxical argument to that same set? – James Arathoon Dec 13 '17 at 19:57
  • To clarify where I write choose above, you should also read "could choose" for the logic of possibility. – James Arathoon Dec 13 '17 at 20:52

2 Answers2

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Distinguish "I can buy-a-soda-and-a-candy-bar" from "I can buy a soda and I can buy a candy bar".

With $A$ for "I have two dollars", $B$ for "I can buy a soda" and $C$ for , "I can buy a candy bar",

$$((A \to B) \land (A \to C)) \to (A \to (B \land C))$$

says "Given if I have two dollars, I can buy a soda, and if I have two dollars, then I can buy a candy bar, then if I have two dollars, I can buy a soda and I can buy a candy bar”. Which is a tautology, and intuitively fine (if you have two dollars, you can buy the soda, and you can buy a candy bar -- nothing is said about your ability to buy both at the same time!).

But that wff doesn't transate the sentence you gave on the natural reading, as the natural reading of "I can buy a soda and a candy bar" (hooray, I can get both my treats!) is not rendered by $(B \land C)$.

Peter Smith
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  • A tautology is for example a valid mathematical deductive proof where the assumptions used imply the deductive result. “If I have two dollars, I can buy a soda, and if I have two dollars, then I can buy a candy bar, so if I have four dollars, I can buy a soda and a candy bar” is a tautology because I am effectively saying the same thing twice, and if I negate my inference I end up with a contradiction. – James Arathoon Dec 13 '17 at 16:47
  • Ignoring of course the distinction between buy and afford and whether or not the shop keeper will sell both products to a single buyer. – James Arathoon Dec 13 '17 at 16:50
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“If I have two dollars, I can buy a soda, and if I have two dollars, then I can buy a candy bar, so if I have two dollars, I can buy a soda and a candy bar”

I have come up with the following: ((A → B) ∧ (A → C)) → (A → (B ∧ C))

Well, the argument itself is unsound, as you could not simultaneously spend the same two-dollars on different purchases, but anyway, yes, that is an accurate rendition of the statement you were told to translate.

So, to prove that the statement is a tautology we just note informally that : If it is true that both $(A\to B)$ and $(A\to C)$ hold simultaneously, then it would be so that if $A$ held then both $B$ and $C$ would hold.  

$\\[12ex]$

A sounder argument would be: "I may buy a soda if I have two dollars, or I may buy a candy bar if I have two dollars, so I may buy a soda or a candy bar if I have two dollars."

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Graham Kemp
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