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I just started studying logic (high school) anyway...for the truth table of logical implication

If sentence $A$ is true and $B$ is true then $A\implies B$ is true.

does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?

the same for if $A$ is false can you get anything either True or false proved from this $A$?

Hanul Jeon
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lina
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  • If $A$ is true and $B$ is true, then you can't simply say $A\implies B$. – gebruiker Sep 27 '14 at 10:09
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    @gebruiker Yes, you can. – Git Gud Sep 27 '14 at 10:10
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    @gebruiker - Why would you think you can't ? – Belgi Sep 27 '14 at 10:20
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    @Belgi Because I own a dog is true, and I am a man is true. But owning a dog does not imply being a man. I know women who own dogs too. Or am I overlooking something here? – gebruiker Sep 27 '14 at 10:32
  • A few remarks about this question. The way the question is phrased and the fact that the OP is a high school student makes me believe that in this context there is no distinction between syntactic truth and semantical truth. (In most deductive systems in classical logic, in a sense, there is in fact no difference between them due to the completeness theorem for propositional calculus, see this recent question). – Git Gud Sep 27 '14 at 10:43
  • So, in my opinion, a good answer would alert to the existence of this two views, not necessarily explaining everything, but at least explaining what one means with prove, deduction, etc as this terms are being used loosely on this whole thread. – Git Gud Sep 27 '14 at 10:45
  • @gebruiker You are overlooking something. Two things. For one, I believe you're unaware of what a (formal) proof actually is, (see my comment above). The other one is that your example isn't adequate for this situation, it's not even propositional calculus. The statement gebruiker is a man and gebruiker owns a dog can be formalized in propositional calculus, but there exist women who owns dogs cannot. – Git Gud Sep 27 '14 at 10:52
  • This can be fixed by considering a specific woman you know, let's say her name is Anna. Then you can say Anna (who is a woman) owns a dog, but how is this supposed to show that there isn't a proof of if gebruiker is a man, then he owns a dog? – Git Gud Sep 27 '14 at 10:53
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    @gebruiker "(gebruiker owns a dog) implies (gebruiker is a man)" is true. "(X owns a dog) implies (X is a man)" is not true for all X – Ben Aaronson Sep 27 '14 at 20:33
  • gebruiker owns a dog and gebruiker is a man, therefore gebruiker owns a dog $\Rightarrow$ gebruiker is a man. This is not to say that everyone who owns a dog is a man; that's a whole other statement. – Theodore Norvell Sep 27 '14 at 20:36
  • You say "if sentence A is true and sentence B is true then the sentence A=>B is true; does this mean that there is a way to prove that B is true from A?" But you have already stated as a condition that B is known to be true, so you don't need a proof of B; B is already known to be true by your assumption "... and sentence B is true..." – Eric Lippert Sep 27 '14 at 23:02
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    $A\implies B ;;;;$ means $;;; \operatorname{truthvalue}(A) \leq \operatorname{truthvalue}(B) ;;;$. $;;;;;;;$ –  Sep 28 '14 at 02:01
  • @gebruiker Correlation does not imply causation. – Cole Tobin Sep 28 '14 at 05:50
  • @RickyDemer That's a completely unhelpful way of looking at it because it assumes that truth values are numbers or are ordered according to some standard order. They're not. For example, it's quite common to treat the domain of a Boolean function as being ${-1,1}$, where $1$ corresponds to false and $-1$ to true. – David Richerby Sep 28 '14 at 09:08
  • @ColeJohnson I guess the point here is that correlation does actually imply causation, IF the correlation is perfect. If 'all persons are men' and 'all persons own a dog', then that does imply that 'being a man implies owning a dog' (and vice versa). – Thomas Ahle Sep 28 '14 at 14:13
  • @ColeJohnson Yes, that's exactly what gebruiker said, it's everyone else who is disagreeing. I agree with gebruiker, too - if you only have one data point and no rules, it is not enough to come up with the rule "A implies B" – Izkata Sep 28 '14 at 16:54
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    @David: Truth values are ordered, since Boolean algebras have a definable order on them. And it's exactly what Ricky wrote. – Asaf Karagila Sep 29 '14 at 16:27
  • @Izkata: gebruiker asserted two specific statements as true. If these statements are $A$ and $B$, then $A\implies B$ as everyone else says. If $A$ and $B$ are instead some propositional formulas, such as "$x$ owns a dog," not the specific instances gebruiker asserted, then neither is known to be true, and nobody claims that $A\implies B$ in that case. The claim that "you can't simply say that $A\implies B$" is based on using two different definitions of $A$ and two different definitions of $B$ within one sentence. – David K Sep 29 '14 at 21:05

11 Answers11

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As a logical proposition, the material conditional $A \implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A \implies B$, for some specific $A$ and $B$, says very little.

Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A \implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.

For example, consider the statement: $$x > 2 \;\land\; x \text{ is prime} \implies x \text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.

Ilmari Karonen
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One way to understand implication is to remember that $A\Rightarrow B$ is equivalent to $\neg A \lor B$. If you understand negation ($\neg$) and disjunction ($\lor$), then you understand implication.

Theodore Norvell
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Look at $A$ and $B$ as something that is either false or true. For example let $A$ be the event that tomorrow is Tuesday and let $B$ be the event that the day after tomorrow is Wednesday.

Look at $$ A\implies B $$

as a promise - if $A$ is true then so is $B$.

In our example, if $A$ is true then indeed so is $B$ and so the implication $A\implies B$ is true.

However, now consider $C$ as the statement that tomorrow is Friday, and I state $$ C\implies B $$

that is - I promise you that if $C$ will happen so will $B$.

Tomorrow is not Friday (at the time of writing), and so $C$ is false, regardless of if $B$ is false or true - my promise was kept.

Now regarding the terminology of proofing $B$ is both $A,B$ are true. Note that statements like $$ \text{My cat walks on four}\implies1+1=2 $$

is true, since both are true, but what would it mean to prove $B$ from $A$ ?

Belgi
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Maybe it's more clear if we separate the logical operator meaning of implication from its logical statement meaning.

When we use it as a logical operator, we conceive it simply as an entity, that given two logical values (thus true or false), produces a third logical value, using a common defined rule (its truth table).
So it makes perfect sense to say $A \implies B = true$ if $A = B = true$, and we don't concern about what actually proposition $A$ and $B$ means, we care exclusively about their logical values.

It's different when we use implication as a logical statement.
In this case we really say something about the meaning of the propositions involved in our statement. So, while proposition $A = My\;cat\;is\;black$ is true, and proposition $B = I\;am\;hungry$ is true as well, $A \implies B$ is not a valid statement.
Such meaning is linked to set theory and formal logic. Using implication in this context means that you can infer $B$ from $A$, in a way called modus ponens.

seldon
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Implication can be understood as thinking in NECESSARY and SUFFICIENT conditions. A good example is to think about someone who has born in Dallas, Texas.

Proposition P could stand for "Someone who was born in Dallas"

Proposition Q could stand for "Someone who is texan"

Truth table for implication is:

enter image description here

So, in this example is easy to check line by line the validity of truth table above:

  • Who was born in Dallas is therefore texan; So the first implication is TRUE.

  • Who was born in Dallas MUST be texan. So, saying that someone who was born in Dallas, in the state of Texas, and is not texan is FALSE;

  • Saying who wasn't born in Dallas but is texan anyway (take someone who has born in Houston, for example) is a TRUE statement;

  • At last, who wasn't born in Dallas AND also is NOT texan is also TRUE.

So, thinking in terms of NECESSARY (to be texan is a necessary condition for who was born in Dallas) and SUFFICIENT (to born in Dallas is a sufficient condition to be texan) conditions ease the process to translate implication to natural language.

Other examples of propositions containing necessary and sufficient conditions are:

  • P = To be a priest / Q = To believe in God;
  • P = To be a Police officer / Q = To carry a gun;
  • P = The number is divisible by 4 / Q = The number is even.
Daniel Lavieri
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I would say that $A$ being true and $B$ being true does not mean you can always prove (deduce) $B$ from $A$.

Here's an example. A: Alice lives in Atlanta. B: Bob lives in Boston. Even if these are both true, there is no (apparent) relationship. So you can't logically deduce $B$ from $A$ even though $A\Rightarrow B$ is true in this case.

I guess this means that if you can logically deduce statement $Q$ from statement $P$, then $P\Rightarrow Q$ is true; but knowing $P\Rightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption of $P$.

Logical implication is a defined logical connective, so as long as $P$ and $Q$ have truth values (true or false), so does $P\Rightarrow Q$.

paw88789
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  • You can logically deduce B from A in that example. Let's assume A is true. B is true. QED. Yes, this counts, unless you really meant A implies B and also not not-A implies not-B. But if you meant that, you should have said so. – djechlin Sep 28 '14 at 02:00
  • Your 3rd paragraph is incorrect. Suppose $P\Rightarrow Q$ is given, we construct a proof of $Q$ to $P$:

    Assume $P$, assume $\neg Q$, then by the truth table for $P\Rightarrow Q$ (or by modes ponens) $Q$. Hence $Q$ and $\neg Q$, a contradiction.

    Hence our assumption $\neg Q$ was false.

     – James Sep 29 '14 at 16:41
  • My point is that sometimes two statements, $P$ and $Q$ are unrelated. Now $P\Rightarrow Q$ can perfectly well be true (provided $P$ is false or $Q$ is true), and yet knowing that $P$ is true does not help us determine (deduce) that $Q$ is true, because the statements are unrelated. – paw88789 Sep 29 '14 at 17:43
  • @paw88789 But when you know $P$, then to quote you "knowing $P\Rightarrow Q$ is true does not guarantee the existence of a deduction of $Q$ from the assumption that $P$." This is wrong, I just gave a deduction. – James Sep 29 '14 at 18:42
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The problem is understanding that there's a difference between the everyday mathematical meaning of "implies" - "I can prove it (or someone can)" - and the formal logical meaning expressed by the truth table. This may help: In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True?.

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Ethan Bolker
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This is the way I understand it (I make no guarantee this is considered a correct interpretation).

the truth value of an implies statement is basically 'innocent until proven guilty', or 'my statement is true so long as the proposition values don't disprove it'.

So with an example, propositions P and Q, with the statement P implies Q. Lets say proposition P is 'is a man' and proposition Q is 'lives in japan', so our implication statement is basically saying 'all men live in japan'.

The truth table:

truth table

is basically saying, for any example where P and Q are true, we haven't disproved our statement that P -> Q, and for any example where P is false we haven't disproved our statement, only if we have an example of P being true and Q being false is our statement shown to be false.

So imagine there are only 4 humans in existence, we look at each of them and categorise them in our truth table. By doing this we can work out if our statement that P -> Q is true. We find 2 of the people are men, and that they do live in Japan, OK good, our statement still holds true, the 3rd person is a woman, we don't care if she lives in Japan or not because in either case our statement about men is still holding true.

A woman living in or outside japan has no bearing on the truth of our statement about men living in japan.

The 4th person then will decide if our statement is true or not, if the 4th person is a man living in japan, or a woman, then is it logically true to say that being a man implies that you live in japan.

Why is this useful? well the truth table itself doesn't help us prove the truth of our statement by itself, what it does is clearly set out what conditions are relevant to the truth of our statement - i.e. that we don't care about women, and that only a man not living in japan will falsify our statement.

Of course, you could make a statement like 'all green men live on mars'. This is a true statement because we cannot verify who lives on mars, and we don't have any examples of green men. But this is whats called a "vacuous truth", it has no bearing on reality for us because the propositions are intangible.

Coder375
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If sentence $A$ is true and $B$ is true then $A\implies B$ is true.

Yes. In mathematics and logic as widely applied, $A\implies B$ is equivalent to $\neg[A\land\neg B]$.

So, if $A$ is true and $B$ is true, then $\neg[A\land\neg B]$ and hence $A\implies B$ will also be true.

Does that mean if $A$ and $B$ are both true then there is a way to prove $B$ is true from $A$, always?

Yes. Here is a formal proof using a form of natural deduction:

enter image description here

The same for if $A$ is false can you get anything either True or false proved from this $A$?

Yes. Here is a formal proof:

enter image description here

That the above may seem counter-intuitive may stem from the fact that implication is often confused with temporal notions of causality. The statement, "If it is raining, then it is cloudy" does not mean that rain always causes cloudiness, or that cloudiness always cause rain. It means only that, at a given instant in time (usually the present), its is not the case that it is both raining and not cloudy.

Dan Christensen
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  • This is too complicated. Why don't you just take a basic statement and elaborate. For example there is a norm for males of running 100m in less than 10.05 seconds to qualify for the Olympic games. https://pbs.twimg.com/media/D1VNW3eUYAAvwsn?format=jpg&name=large

    Lets make P = "one runs 100m in less than 10.05" and Q = "one does qualify for the Olympics". What happens then?

     – Cornelius Nov 06 '20 at 14:14
  • Until students understand some of the most basic methods of proof (as in my first proof of $A \land B \implies (A\implies B)$ above), they will simply have to apply the definition $A\implies B\equiv \neg(A \land \neg B)$. – Dan Christensen Nov 07 '20 at 03:55
  • I've made you a most simple example, and you can't make it work? – Cornelius Nov 08 '20 at 09:57
  • @Cornelius As for your simple example, "What happens then?" is a bit ambiguous. You have, of course, $P\implies Q$. But that is not something that "happens." It is simply not the case that both P is true and Q is false. – Dan Christensen Nov 08 '20 at 18:04
  • So, if P is false, and Q is true, the outcome would be "you didn't run 100m under 10.05 => you have qualified for the olypmics" which is not true. So ⊥=> ⊤ is ⊥? – Cornelius Nov 13 '20 at 00:11
  • @Cornelius As I have proven above, if $P$ is false, then, for ANY logical proposition $Q$ (be it true or false), the implication $P \implies Q$ is always true. That doesn't, however, mean the $Q$ must be true. Neither does it mean that $Q$ must be false. – Dan Christensen Nov 14 '20 at 01:38
  • New versions of proofs posted 6 years later. – Dan Christensen Nov 14 '20 at 03:49
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The sentence $A\implies B$ is not a sentence about proving anything. It is a compound sentence, but otherwise just like the sentences $A$ and $B$. For the truth table, the compound sentence got to have truth values like $A$ and $B$ has, and it has to depend on these sentences truth values. The values for this compound sentence is the same as for the compound statement $\neg A \vee B$.

Lehs
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This illustration of logical implication might help:

a) When you truly understand logical implication,
b) then you’ll be a happy person.

You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication. :-)

Truth table:
a b a⇒b
F F T
F T T
T F F
T T T

I hope this somewhat intuitive example helps.

Bert Sierra
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