Intuitively, why does (B → S) ≡ (¬B ∨ S)?

Question

I pursue only intuition; please do not answer with formal proofs or Truth Tables. I comprehend the following explanation, but it still does not supply the intuition and I sense a deeper, directer, more intuitive explanation. So how else can I intuit this? Or have I neglected something below?

_{I rewrote: p 51-52. Sweet Reason: A Field Guide to Modern Logic (2010 2 ed) by Henle, Garfield, Tymoczko.}

2. If the kid misbehaves and the teacher forces him to stay, then the teacher spoke truthfully.
3. Suppose the kid misbehaves but travels. Then clearly the teacher lied.
4. But what if the kid misbehaves, but a big snowstorm forces the trip’s cancellation? You can’t say the teacher lied. The kid behaved but didn’t go; nobody went. The teacher spoke truthfully. 1. If the kid behaved and travels, the teacher spoke truthfully.

Notice that the pattern of Ts and Fs we get, is the pattern for disjunction (see p. 38).

Answer 1

It might be helpful to think of this by considering each of the disjuncts of a disjunctive statement as containing unspoken implications. Consider, for example, the following disjunction:

Either no storm will come, or the boat will sink.

For each of the disjuncts, we can assert the following implications:

• If no storm comes, no storm comes.
• If a storm comes, the boat will sink.

It's intuitively true that either a storm will come or not. Therefore, we can put all that together to derive the following, which is a logical consequence to the first statement:

If no storm comes, no storm comes; or if a storm does come, the boat will sink.

The first part is redundant, so this can be simplified to:

Either no storm comes; or if a storm comes, the boat will sink.

From this it can be seen that disjunctive statements can be thought of as having conditional statements hidden within their elements.

Vacuity

It might be argued that it is vacuous to insert an antecedent into the disjunction in that way. For example, it could be argued that any antecedent could be inserted, such as:

Either no storm comes; or if the moon is made of cheese, the boat will sink.

Although it is true that that logically follows from the premise, there's an important difference to be noted. The antecedent "if a storm comes" complements the other disjunct, whereas "if the moon is made of cheese" does not. This involves the important difference that the simpler conditional follows logically from the disjunctive statement:

• ~A ∨ (A → B) implies A → B
• But, ~A ∨ (C → B) does not imply C → B

In other words, it cannot be concluded that if the moon is made of cheese, the boat will sink; but, it can be concluded that if a storm does come, the boat will sink.

Edit:

Due to a question in a comment, I'm adding the following proof:

{1}      1.   ~A ∨ (A → B)                Prem.
{2}      2.   A                           Assum.
{3}      3.   ~A                          Assum. (1st D)
{4}      4.   ~B                          Assum.
{3,4}    5.   ~A & ~B                     3,4 &I
{3,4}    6.   ~A                          5 &E
{2,3,4}  7.   A & ~A                      2,6 &I
{2,3}    8.   B                           4,7 RAA (1st C)                       
{9}      9.   A → B                       Assum. (2nd D)
{2,9}    10.  B                           2,9 MP (2nd C)
{1,2}    11.  B                           1,3,8,9,10 ∨E
{1}      12.  A → B                       2,11 CP    

The best way to prove an implication is usually to start by assuming the antecedent of the implication to be concluded (line 2). To eliminate the disjunction, I also have to assume each of the disjuncts (lines 3 and 9). The first disjunct contradicts the assumption in line 2, and given a contradiction, you can essentially conclude whatever you want. I needed to conclude B, but it should be noted that I didn't get that conclusion for free; it depends on the assumptions of lines 2 and 3 which will have to be eliminated (line 8). The second disjunct also lets me conclude B using modus ponens. Since both disjuncts imply the same conclusion of B (line 11), I can eliminate the assumptions of lines 3 and 9. The assumption on line 2 led to the conclusion on line 11, so I can eliminate it and conclude that A → B.

Answer 2

Clearly, it is not the case that just because two things are true, the one logically implies the other in any realistic way. Nor is it true that whenever something is false, it logically entails things that are not remotely related to it.

If something is simply not really true, in a realistic sense, it is not going to be intuitive. The material conditional is not defined as it is because it captures an intuition, it is defined as it is because it is safe to extend the realistic definition this far without wandering into dangerous territory.

So we really have to trim the question down to this: Why, intuitively, is such overstatement safe?

If something really does logically imply something else, say A |- B, then clearly whenever A is true B is forced to be true. And B might be true anyway, for some reason other than A. So whenever B is true, A can be true or false. The logical connective that is always true whenever one of its arguments is true is OR. So all the cases we really care about are in fact surrounded by B v ~A, although when A or B are not really logically connected, we are making up meaningless truth values, just to have a truth value.

So, if we take this overstatement seriously, does it cause problems?

Well, if A and B are not really logically connected and A is a true premise, there may be some incredibly indirect way that we cannot see in which they will eventually be seen to be connected. We do not want to block those potential paths. If we do we will be creating an artificial opportunity for an incorrect reductio-ad-absurdum argument. Letting all true premises imply one another avoids being tripped up by our lack of insight.

Similarly, if A is a false premise, we should not be deducing anything from it, but if there is anything we cannot deduce from it, we don't want an unexpected contradiction, because we want to trust reductio-ad-absurdum as a technique. So just letting it imply anything and everything prevents an illogical proof-by-contradiction from falling out of the sky.

Between these two observations, it is clear that (somewhat amazingly) our abstraction fills in the surrounding truth values in the safest, most conservative possible way, and avoids illogical contradictions when they might slip in.

Answer 3

B → S is more intuitive; most people more or less have the feeling that if you say this then you also mean that without any intervention of "not" and "or"; this is the origin of implication.

B → S only mandates that it cannot be the case that B is true and S is false, i.e. ¬(B.¬S) (See PM ✳1 Primitive Ideas, Definition of Implication). Thus ¬B ∨ S is just a logical convenience that expresses the same connection between B and S; there is probably no "intuition" behind the equivalence of the two. The reason we continue to use → is that it is intuitive and directly express the connection between B and S without the intervening ~B.

Same kind of thing can be said about Sheffer stroke: "or" expresses hesitation; "not" expresses rejection; both can be expressed in terms of Sheffer stroke. We concede that Sheffer stroke is more fundamental, but I doubt anyone can "intuit" the relation symbolized by this vertical bar. "or" and "not" exist in many natural languages' minimum vocabulary, "nand", as far as I know, does not - please enlighten me if you know a language that has "nand" in it.

Answer 4

I sense a deeper, more direct intuition

For sure intuition was required to formulate the logical calculus formally; and it's this intuition that is relied upon when this calculus is first described, in the same way that when we're introduced to arithmetic we're relying on how we understand actual physical arrangements of buttons or bottles.

It's this intuition that is pressed into service for philosophical as opposed to mathematical or formal logic, it asks the kind of questions that you're asking here.

Still, it ought to be noted that there is a line of thought that attempts formalisation arguing that it makes for efficient symbolic manipulation - usually this is at some cost or trade-off which may include some loss of intuitive clarity at certain places of the formalisation.

Answer 5

When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S. I prove both directions of this equivalence.

Proof of (→): Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional - the only way it can be falsified is if S does not hold (this is logical implication, not causation). Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless - how could we say it were false? Therefore, B holds.

Proof of (←): Suppose that B ∧ ¬S. Then B is true, but S is false. So the conditional B → S is false.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.

Answer 6

That is easy. B → S means (nothing else but) either you have S or not B. This is obvious once you consider

1. the banned case, where S is negative,

2. because then having B positive is impossible since B, implies S and thus contradicts 1.

You cannot have B and ¬S at the same time. We therefore ban (B and ¬S) and are left with ¬(B and ¬S), which ≡ ¬B ∨ S.

Answer 7

I tend to think of such things as fluent state diagrams in simple cases like this (less the transitions). Take the diagram below, for example, where we enumerate the possible combinations of the fluents B and S:

Thinking of B → S and (¬ B ∨ S) separately would lead me to eliminate the same states (1 and 2) in the diagram. Note that I am treating the disjunction as exclusive, hence eliminating 1.

Since the same states are eliminated when considering LHS and RHS I conclude that, indeed, B → S ≡ (¬ B ∨ S).

Answer 8

I have just been reminded of another good reason why this should not be intuitive: It results in Curry's paradox.

Curry's paradox is: If this sentence is true, then all sentences are true.

But of course not all sentences are true.

So Curry's sentence is not true.

But if falsehood implies everything, since the left-hand side of the implication is false, the sentence is true.

Nothing directly equivalent to a paradox should be considered intuitively obvious, only naive.