If both $P$ and $Q$ are true , how can I tell that $P$ implies $Q$?

Question

I am trying to understand the fundamentals of mathematical logic, in order to be able to study discrete mathematics and computer science, and have a big problem understanding Implication.

Intuitively, I understand very well the implications $$ x >0 \implies 2x>0\\ x \text{ is a prime number} \implies x \geq 2,$$ because in each case there is a connection between the hypothesis and the conclusion; they are related, and fall in the same context. However, any two propositions, even when they are not related at all or belong to different contexts, can be linked through an implication. For example, according to their truth tables, the implications $$ \text{A day is 24 hours} \implies \text{A cat has four legs and a tail}\\ 2 \text{ is a prime number} \implies \text{An hour is 60 minutes}$$ are logically and mathematically True, since in each case both inputs are true. My first question is, how can this be??

My second question is related: How can we use truth tables with implication anyway? I understand that truth tables are used to list the possibilities of the output based on the logical values of the inputs; in any line of the truth table, the value of the output depends only on the logical values of the inputs, and has nothing to do with the conditional connection between the two inputs. How can we display a conditional statement in a truth table the same way we display a logic gate or so? In other words, how can I tell only from the values of $P$ and $Q$ that $P \implies Q$?

Answer 1

Here is the truth table for an implication:

$$ \begin{array}{ccccc} P & & Q & & P \to Q \\ T & & T & & T \\ T & & F & & F \\ F & & T & & T \\ F & & F & & T \end{array} $$

You can think of an implication as a conditional promise. If you keep the promise, it's true. If you break the promise, it's false.

If I tell my kids, "I'll give you a cookie if you clean up." Then they clean up. I better give them a cookie. If I don't, I've lied. However, if they don't clean up, I can either give them a cookie or not. I didn't promise either if they didn't keep up their end of the bargain.

So in other words, an implication is false only if the hypothesis is true and conclusion is false.

Logically $P \to Q$ is equivalent to $\neg P \vee Q$

I had a computer science professor who was fond of promising his kids things given a false premise. This way he wasn't compelled to follow through. Example: "If the moon is made of green cheese, I'll give you an x-box."

Answer 2

Think about it this way:

If a day is 24 hours, does a cat have 4 legs and a tail?

Even though they are unrelated, the answer is yes, so a 24-hour day implies that a cat has 4 legs & tail.

If 2 is a prime number, does an hour have 60 minutes? Yes.

Going with the cookie idea, let's say you already have a cookie in your hand and you are about to give it to your kid. She says "If I clean my room, can I have a cookie?" You will likely say "yes!" Technically, it's true that you'll give her a cookie if she cleans her room. She doesn't need to know she was going to get it anyway.

Answer 3

As long as all atomic propositions P, Q, R, ... within a compound formula C get assigned truth values in {0, 1}, it holds that C will take on a truth value in {0, 1}. Here 0 indicates falsity and 1 truth. By atomic proposition, I mean any variable such a "p", "q", or "r" or constant such as "1" or "0". This consists of a way of stating the principle of truth-functionality in classical propositional logic. So, as long as p and q get assigned truth values in {0, 1}, (p->q) will have a truth value in {0, 1} by this principle. Here "->" indicates a truth function of two arguments, specifically the material conditional.

So, if you assign both p and q a truth value of 1, then just by the principle of truth-functionality (p->q) will have a value in {0, 1}. So, if p is true, and q is true, is (p->q) false or true? Well, you pointed out above, (x >0) ====> (2x>0) is true, when (x>0) is true and (2x>0) is true. So, you know of at least one instance with (p->q) true with both p true and q true as well. Now, if you were to accept a single instance of (p->q) as false with both p true and q true as well, then you would either have a violation of the principle of truth-functionality in classical propositional logic, or you would render it inconsistent (since (p->q) takes on truth value true in one case where p and q come as true, and another where (p->q) takes on truth value false in another case where p and q come as true). Both come as problematic for classical propositional logic. So, basically to speak consistently you have to accept (p->q) as true in classical propositional logic anytime you have p true and q true.

The catch lies in that the truth values of the atomic propositions (the variables or constants) and the definitions of the logical operations (truth-functions) completely determine the truth value of a particular instance of something like (p->q). The meaning of the "p" and "q" bears utterly no relevance on the truth value of (p->q), only the truth values of "p" and "q" do. The principle of truth-functionality entails this.

Now, there do exist other ways to interpret a word like "implies" other than by "->", or some other symbol which might get used for the material conditional. As one well-known example, some logicians have interpreted "implies" as a strict conditional. However, these interpretations refer to something which is not truth-functional. To reiterate, a true statement A can "imply" a seemingly unrelated statement B, because of the principle of truth-functionality. The meaning of something like (p->q) isn't so much that p implies q in the sense of some sort of connection between p and q, but rather that it is not the case that p stands as true and q stands as false simultaneously.

Truth tables don't list probabilities. They tell you what logical value of the output IS on the basis of the input(s). The truth table depends on the values of the inputs, AND the function, that is the connective, which links them together. The truth tables for conjunction and the material conditional have the same inputs, however, they have different connectives linking together the variables, and thus have different outputs in their final column.

As this truth table indicates:

p   q  (p->q)
0   0     1
0   1     1
1   0     0
1   1     1

If p is false, then (p->q) is true. If q is true, then (p->q) is true. If p is true, then (p->q) has the same truth value as that of q. If q is false, then (p->q) has the same truth value as the negation of p.

Addendum: In classical logic any given true statement (tautology) comes as related to any other true statement and belongs to the same context as any other true statement. They come as related in that given any true statement, you can derive any other true statement. They belong to the same context in that they exist within the scope of a single derivation.

Answer 4

Yes, implication like this is weird. You feel like the antecedent must be somehow relevant to the consequent. It just turns out that it makes mathematical reasoning much ...cleaner... if you operate with it using the truth table rules.

The way you should think of implication in mathematical setting is by how upset you would be if some one claimed 'P implies Q' based on which of P and Q are true or not. The only situation where you should be upset is if P is true and Q is false (that definitely goes against what implication means). If P is true and Q is true, then obviously implication works out. What if P is false? Should you be mad/upset whatever Q is? I don't think so.

Convince yourself that it should be OK. And anyway mathematical culture has accepted that such a (stipulated) definition (of implication as only being false when P is true and Q is false) is the most useful way to define it.

This is really just a wordy explanation of the truth table; also an explanation that sometimes things are just defined that way to make things easier. Take for example the natural number 1. Is it a prime? by most informal definitions of 'prime' it is. It just turns out that number theory theorems and proofs are simpler to state if you treat 1 separately (and call it 'unit').

Answer 5

Intuitively, I understand very well the implications $$ x >0 \implies 2x>0\\ x \text{ is a prime number} \implies x \geq 2.$$ However, any two propositions, even when they are not related at all or belong to different contexts, can be linked through an implication. For example, $$ \text{A day is 24 hours} \implies \text{A cat has four legs and a tail}\\ 2 \text{ is a prime number} \implies \text{An hour is 60 minutes}.$$ How can this be??

1. Reading the implication $$\text{if }P\text{ then }Q\\P\implies Q\tag1$$ literally, without projecting connotations beyond logical deduction, isn't $P$ literally just a hypothetical and $Q$ the corresponding deduction? (To be clear: the connective "implies that" does not in itself indicate causality/chronology at all; here is just a Sherlock-Holmes process of evidence-based logical deduction.) As such, when its hypothetical $P$ is satisfied, then surely the conditional statement is true precisely when its asserted deduction $Q$ is true?

   As long as we are not discussing modality or counterfactuals (where $P$ and $Q$ might stand for assertions like "I will marry you" or "I would have died"), then to assert $(1)$ is to assert that $$P \text{ is sufficient to deduce }Q\\P \text{ is evidence of }Q$$ rather than that $$P \text{ results in }Q.$$

2. Regarding your examples:

   • the propositions "2 is a prime number" and "an hour is 60 minutes" do actually share the same context (and your fourth example implication becomes false when we change the context to Venus)
   • the implication $$x=x\implies\sqrt x\ge0$$ (its quantifier "for every $x$", like for your first two examples, is implicit) contains a hypothetical and a deduction that are evidently "related", so it belongs to your first set of example implications rather than your second set, yet someone who thinks that implications are about cause and effect would still find this one disconcerting.

Answer 6

Our common thinking of implication ($P \implies Q$) is that P allows us to say wether Q is true. But if we already know that Q is true, this way of thinking does not make sense anymore.

Your confusion stems from the fact that you think P would have any implication on Q, which cannot be the case because Q is already given. The statement is true nevertheless, see here.

Take the example given in an other answer, where "clean up" implies "get cookie". I think it helps understanding a special case, but it is not a good one, because we can assume "clean up" would lead to "get cookie", if "get cookie" were false. But the statement holds true even for things that will never lead to "get cookie". In other words, if it is really decided, that "get cookie" is true, you can say anything in your if-sentence and the statement still holds:

If you crash the car, I will give you a cookie.

Our common understanding of this sentence is, that the cookie is the reward for crashing the car. But this does not make sense in this situation, so nobody would normally formulate it this way, but rather:

You can do anything, I will give you a cookie.