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Having studied logic, I still cannot understand the conditional. At first, it was because (as with most things I learn) it was a problem with my understanding. I now believe it is because there is an actual problem not properly understood.

So, as you know, the conditional assesses the statement 'If A then B'. Now, I strongly presume that this means if A is true, because in a way you could say if A is false you're still considering the value of A, so that even if A is false it still means 'if A'. Though from what i've read the statement only holds if A is true. After All, surely 'if A' ≠ 'if not A'.

Ok sweet, lets right down the truth table:

A    B   Value
______    
1    1:    T
1    0:    F
0    1:
0    0:

So far i've only done half the truth table. That half is within the bounds that we know that A is true.

Why haven't I completed the other half? Well, the truth value refers to the statement 'if A then B', we do not know what happens when A does not occur.

To resolve this, logicians have chosen to set the value of the proposition to be true for when A is not true by default. Now, it can't be the case that all factors are held constant (except A, of course), for when A is false, otherwise it would literally be impossible for B to be occur and not occur given the exact same conditions.

So, by deduction, we have to consider the fact that the state of the world is different for when where A is false yet B is true, and for when A is still false yet B is also false.

In doing so, this means that if A is false yet B is still true, then something tells us that there are other causes which lead to B. Once again however, the condition is 'if A then B', not 'if A and possibly some other factor then B', or, where X is some arbitrary factor, 'if A or X then B'= '(A+X) -> B'.

So the main problem I have here, is that we are trying to assign a truth value for other conditions that we initially set ourselves, which I think is bad logic. I strongly believe we should just consider the truth value for where A is false to be unknown.

If we don't, then would we not have to accept that what we're really saying is by 'If A, then B' is actually 'If A or some other unknown factor X, then B'?

The issue in the conditional crosses over to other areas as well. For e.g, take the case 'A unless B' = ~B -> A = A V B.

However, if B occurs, and A still occurs despite the fact B still occurs, then this is still a true statement. This means that 'A unless B' is NOT saying that A won't occur even IF B occurs.

user2901512
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2 Answers2

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Material implication is good for statements like, "if $x$ is rational, then $x^2$ is rational" -- which is a statement that we would like to think of as true, and which we logically render as $$ \forall x\in \mathbb R(x \in \mathbb Q \to x^2 \in \mathbb Q). $$ We want the inner statement to be true for any $x$. In particular, we want it to be true for $\sqrt 2$, where the antecedent is false while the consequent is true. Edit: and we want it to be true for $\sqrt[4]2$, where both the antecedent and the consequent are false.

Mees de Vries
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  • If A = 2(1/2) and B = 2, then saying that A = 0 and B = 1, is true, yes. However, A = 0 and B = 0, is false, though interpreted as also true. How can that be right? It's like am the only one that has a problem with that :/ – user2901512 May 17 '16 at 15:38
  • You agree that my sentence should be true, right? So the part in parentheses should be true for all $x$? In particular, it should be true for $x=\sqrt[4]2$? Which of these statements do you disagree with? – Mees de Vries May 17 '16 at 15:52
  • Well, if x is x is 2^(1/2), then saying that 2(1/2)^2 does not lead to a rational number, is false. However, isn't this interpreted as true given the conditional statement? – user2901512 May 17 '16 at 15:59
  • I am talking about the quartic root of 2, not the square root. – Mees de Vries May 17 '16 at 16:02
  • But if your talking about the quadratic root, then setting A to false and B true is not correct either. The thing is, is that x^2, no matter what it is, can't be both rational and irrational. So where A is false, the statement x^2 is rational (B is true) and x^2 is not rational (B is false) can't BOTH be true. Something has to give. – user2901512 May 17 '16 at 16:07
  • I'm afraid I don't understand what you're asking. – Mees de Vries May 17 '16 at 16:16
  • Well, the statement 'if x is rational, then x^2 is rational' follows the logical form if A thus B. So if both A and B are false, this is interpreted as a true statement. Lets say x is 2(1/2) = sqrt(2). Saying that both A and B are false, is saying x^(1/2) is irrational, and that 2^(1/2)^2 is also irrational, which is false, but yet still stands as a logically correct argument. – user2901512 May 17 '16 at 16:19
  • If $A$ and $B$ are false, then indeed $A \to B$ is true. However, $x = \sqrt 2$ in the above sentence is not an example of that. Because then $B$, which says $x^2 = 2 \in \mathbb Q$ is true. However, if $x = \sqrt[4]2$, then $A$, which says $x \in \mathbb Q$, is false, and $B$, which says $x^2 = \sqrt2 \in \mathbb Q$, is also false. And we want $A \to B$ to be true in order for the entire sentence $\forall x \in \mathbb R(x \in \mathbb Q \to x^2 \in \mathbb Q)$ to be true. Thus, false $\to$ false should be true. – Mees de Vries May 17 '16 at 16:58
1

According to your table, (P -> P) is not a tautology, because the fourth row is left blank. It doesn't make much sense to say that (P -> P) is not a tautology.

According to your table also, (P -> (Q -> P)) is not a tautology. Given that (P -> P) is a tautology, (0 -> 0) = 1. So, (0 -> (0 -> 0)) = (0 -> 1) which remains indeterminate and thus (P -> (Q -> P)) is not a tautology. If (P -> (Q -> P)) is not a tautology, then you can't start with an assumption A, derive a conclusion C, and then infer that (A -> C), because you can only do that when a system either assumes the rule of conditional introduction (which I more-or-less just said) or has enough to prove the deduction meta-theorem which implies that (P -> (Q -> P)) is a tautology (so long as the system is sound). If (0 -> 1) = 0, then (P -> (Q -> P)) is not a tautology. That again, implies that you can't start with an assumption A, derive a conclusion C, and then infer that (A -> C). Consequently, so long as you want to have the ability to start with an assumption A, derive a conclusion C, and then infer that (A -> C), it holds that (0 -> 1) = 1, and thus the table for the material conditional holds.

So, no, the truth table for the material conditional should remain as it stands.

Doug Spoonwood
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  • I don't see what you by (P -> P) is not a tautology, though. If you have one variable then wouldn't it just be a 2 bit truth table? For a conditional, wouldn't you need to consider the effect one variable has on another? – user2901512 May 17 '16 at 22:49
  • @user108262 No, you don't consider the effects of one variable on another, because variables don't have effects on each other. The truth table for (P -> P) is two lines, yes. It entails that (0 -> 0) = 1 and (1 -> 1) = 1. – Doug Spoonwood May 17 '16 at 22:53
  • Oh right sorry, I mean propositions. And yes, it would imply 0 -> 0 = 1, but only it also implies 0 ->0 = 0, which is impossible, if, 0 -> 0 =1. But of course 0 -> 0 must = 1, it's just that, when those truth values are assigned to propositions, 0 would mean 'if not P', but we only know what happens 'if P' occurs, right? – user2901512 May 17 '16 at 22:55
  • @user108262 Propositions don't have effects. Propositions consist of statements. Statements assert, deny, describe, etc. They do not act, and thus cannot cause anything, and consequently don't have effects. – Doug Spoonwood May 17 '16 at 22:58
  • Yes I know, my mistake. I did mean propositions though. That is, when I said variables I had the idea of a proposition in my mind. But yes, variable was definitely the wrong term. – user2901512 May 17 '16 at 22:59