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If you were asked to explain that p -> q is logically equivalent to ¬p v q to someone not engaged in the study of logic, how would you do?

The classic explanations I know are:

1) with truth tables;

2) p -> q holds when p is false or q is true, so ¬p v q.

Do you happen to know any other explanation that can be phrased in layman's terms?

  • I think one has to be engaged somewhat with the study of formal logic. Formal logic and reasoning are not the same thing. Its quite possible for a reasoning layman to not be interested in formal logic though, it happens quite a lot...:) – Mozibur Ullah Jul 30 '13 at 9:58
  • Yes, of course. But let's take it as an academic question: is it possible to explain this to someone not engaged in the study of logic? Your answer seems to be "no", right? – Old Nick Jul 30 '13 at 10:07
  • I don't think so, no. At the very, very least they would need to be interested in formalised reasoning. One would have to explain why this way of formalising reasoning is a good thing, explain what these symbols mean, and how one can apply them, etc. There is a certain amount of commitment thats is required. – Mozibur Ullah Jul 30 '13 at 10:14
  • just as a small clarification, I used the symbolic form (p -> q etc.) but you can state the same equivalence in plain terms, like: "if the sun shines then Carl is fat" is logically equivalent to "the sun does not shine or Carl is fat". So knowledge of symbols is not really required here, I used it just for brevity sake. – Old Nick Jul 30 '13 at 10:56
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I've always liked explaining the material conditional in terms of making promises. Suppose I make the promise "if it gets cold, then I will close the window". I haven't broken my promise in case it either never got cold or I closed the window.

I don't know that this particular example is perfect, but I think the general idea is a good way to sneak in talk of truth-conditions to show equivalence without having to introduce any logical machinery.

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  • I really like this way of approaching the problem, makes sense even for a complete novice. I'll test it asap :) – Old Nick Jul 31 '13 at 7:11
  • It's how I've always introduced the topics in my "baby logic" sections in the first week of classes (for an ethics class, for instance). Seems to work well for those with no background in logic. – Dennis Jul 31 '13 at 8:34
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I believe that most people are confused because in English, "or" is by default the exclusive or. Once you explain that you mean the inclusive or, the rest is pretty obvious. It's not that novices are having trouble understanding how "A or B" relates to the equivalent forms; it's that they think you mean the exclusive or.

Either it's raining or it's sunny. That's how "or" is interpreted in spoken English. (And I assume other languages as well.) When trying to explain things to a novice, start at the beginning.

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As you want to suggest in your comment, you can demonstrate the equaivalence with explicit examples to which it applies, but then this is merely a motivation to define "if-then" in terms of "not-or".

If you start said definition of "if-then", then the semantical explanation 1) shows that you can replace "if-then" with "not-or".

You ask for other elaborations, but if you don't let the semantic explanation count, then to show that equivalence, you must provide the allowed modes of reasoning. This is no quibble, as validity of the rule ((p -> q) -> (¬p v q)) itself is the matter of discourse here.

If we don't want to use semantic explanation, nor formal syntactic deduction for how else "if-then" can be understood, then we'd have to rely on a natural explanation - but you will not be able to find one now, as the "real" reading of "if-then" reasoning was and is a matter of heated philosophical debate. Relevance logic is one approach to it. But p => q always being equivalent to (¬p v q) is not even true in intuitionistic logic. The mathematical structures called Heyting algebras provide a whole family of related formal interpretation of the => connective.

Lastly, I want to comment on your explanation 2). To me it's a little ambigous how to interpret what you want it to say, but either it's a weaker semantical argument than 1) or you say you want to use it to "show" the equivalence deductively. Let's take a look with a classical eye, but without using what we want to show: 2) says ((¬p v q) => (p -> q)) => (¬p v q). Let's investigate the premise (¬p v q) => (p -> q). First level: Assume its premise (¬p v q) and then consider (p -> q). Second level: Assume its premise p. Then classically ¬(¬p) and so by the first level assumption q, so (p -> q) is true. So (¬p v q) => (p -> q) is true anyways and you showed "true => (¬p v q)". From this (¬p v q) actually follows but now the critique of relevance fully applies: The conclusion (¬p v q) wasn't needed in the proof and so you haven't shown much.

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  • I see your point, but if you introduce relevance or intuitionistic logic to explain the equivalence you are over complticating things. My goal is not to introduce a novice into the heart of the philosophical debate about logic, nor to set the correct philosophical background. Its just to make sense of the equivalence in non-technical terms. The (2) in my answer is not to be intended as a "proof", it's just an informal way to explain the equivalence. You can find something similar in Russell PM, most notably. – Old Nick Jul 31 '13 at 7:17
  • @dcmst: You misunderstood me. I mentioned relevance and intuitionistic logic to point out that the equality is by definition. But I think I misunderstood your question too, because from the populariy of the other answers it seems you were just asking for more examles where the equivalence happens to hold. You said "If you were asked to explain that p -> q is logically equivalent to ¬p v q to someone not engaged in the study of logic, how would you do?" To answer this directly: I'd asked why I to explain this, pointing out that the truth of it depends on the form of reasoning you consider. – Nikolaj-K Jul 31 '13 at 8:13

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