I have question about something i don't understand.

$\alpha$, $\beta$, and $\gamma$ are statements.

if $\alpha\implies\beta\lor\gamma$ then it's necessary that $\alpha\implies\beta$ or $\alpha\implies\gamma$. that answer is "no" but i can't understand why.

if it's given that $\alpha\implies\beta\lor\gamma$ then it means that when $\alpha$ is true then $\beta\lor\gamma$ must be also true which means that either $\beta$ is true or $\gamma$ is true. so it means that $\alpha\implies\beta$ or $\alpha\implies\gamma$, isn't it? because the only case when it's might be not true is when $\alpha$ is true, $\beta$ is false and also $\gamma$ is false.

So i can't understand why the answer to this question is "no". I hope someone can explain it.

Thank you. (P.S. sorry for my bad English).

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    $\begingroup$ $\gcd (n,6) \neq 1 \Rightarrow (2\mid n) \lor (3 \mid n)$ $\endgroup$ – Daniel Fischer Jul 17 '13 at 19:43
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    $\begingroup$ Let $\alpha$ be "Propositional logic is boring", $\beta$ be "I am smart" and $\gamma$ be "I am dumb". $\endgroup$ – hot_queen Jul 17 '13 at 19:45
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    $\begingroup$ I believe you're assuming that the "implies" operator is distributive. It is an easy habit to develop because most mathematical operators we deal with have this property. However, not all mathematical operators are distributive. $\endgroup$ – Ataraxia Jul 17 '13 at 19:55
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    $\begingroup$ BTW, did you notice that $(\alpha \Rightarrow (\beta \vee \gamma)) \Rightarrow ((\alpha \Rightarrow \beta) \vee (\alpha \Rightarrow \gamma))$ is a tautology? $\endgroup$ – hot_queen Jul 17 '13 at 20:35
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    $\begingroup$ I don't understand the answers. Shouldn't $P \implies (Q\lor R)$ be the same thing as $(P \implies Q) \lor (P \implies R)$? Their truth tables certainly match. Or am I misinterpreting the question? $\endgroup$ – Alraxite Jul 17 '13 at 22:40

Consider the following statements:

$\alpha$: $n$ is a natural number.

$\beta$: $n$ is even.

$\gamma$: $n$ is odd.

Certainly $\alpha\implies \beta\lor\gamma$, since if $n$ is a natural number, $n$ is even or $n$ is odd.

But we don't have $\alpha \implies \beta$ or $\alpha \implies \gamma$, since not every natural number is even and not every natural number is odd.

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    $\begingroup$ No, this is wrong. This is taking the question as if it is about the relation between $\forall n(\alpha \Rightarrow \beta \lor \gamma)$ and $\forall n(\alpha \Rightarrow \beta)$ and $\forall n(\alpha \Rightarrow \gamma)$, which isn't what was asked. $\endgroup$ – Peter Smith Jul 17 '13 at 20:02
  • $\begingroup$ Peter is correct; this analysis is wrong. While it's correct that the two implications are not true when applied independently, that's not a correct interpretation of the original statement. $\endgroup$ – Dancrumb Jul 17 '13 at 22:45
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    $\begingroup$ @PeterSmith: I appreciate that there are multiple interpretations of the OP's question, but my answer is not "wrong". In first order logic, if we have formulas $\phi(x)$, $\psi_1(x)$, and $\psi_2(x)$, and we can prove $\phi(x)\rightarrow \psi_1(x) \lor \psi_2(x)$, we cannot conclude either of the formulas $\phi(x)\rightarrow \psi_i(x)$. Yes, there are hidden quantifiers here, but I've taken a particular interpretation of the question and provided an example. Based on the wording of the question, I think this will help the OP's understanding more than talking about the material conditional. $\endgroup$ – Alex Kruckman Jul 18 '13 at 0:01
  • $\begingroup$ @AlexKruckman The OP has his Greek letters standing in for statements. Open wffs aren't statements. $\endgroup$ – Peter Smith Jul 18 '13 at 7:02
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    $\begingroup$ @Peter Smith: for the record I have no idea what a "statement" is, and in particular no idea whether it is supposed to be a formula or a sentence. $\endgroup$ – Carl Mummert Jul 18 '13 at 20:28
  1. Put $\alpha = (P \lor Q)$, $\beta = P$, $\gamma = Q$. Then $\alpha \vDash \beta \lor \gamma$, trivially, but neither $\alpha \vDash \beta$ nor $\alpha \vDash \gamma$ (where '$\vDash$' as usual indicates logical entailment, and $P$, $Q$ are contingent). So the principle fails when $\Rightarrow$ is read as $\vDash$, i.e. as logical entailment. (Exactly similarly if it is read as $\vdash$, deducibility in a given proof system.)

  2. Suppose neither $\alpha \supset \beta$ nor $\alpha \supset \gamma$ (where '$\supset$' is the material conditional). Then we'd have to have $\alpha$ true and both $\beta$ and $\gamma$ false, and hence $\alpha \supset \beta \lor \gamma$ would be false too. Contraposing, necessarily, if $\alpha \supset \beta \lor \gamma$ is true, so is one of $\alpha \supset \beta$ and $\alpha \supset \gamma$. So the principle indeed holds when $\Rightarrow$ is read as the material conditional (and the necessity takes wide scope).

So, crucially, you need to be clear about the reading you are supposed to be giving $\Rightarrow$ which (most unfortunately) is used in different ways in different texts.


This answer is closely related to Peter Smith's, but from a somewhat different point of view, trying to show what's common to the examples in the various answers.

Notice first that, as Peter pointed out in the second part of his answer, $\alpha\implies(\beta\lor\gamma)$ is tautologically equivalent to $(\alpha\implies\beta)\lor(\alpha\implies\gamma)$. Things begin to go wrong when these formulas are put into a context that explicitly or implicitly involves a universal quantifier and one tries to distribute this quantifier across the $\lor$ connective.

Thus, in Alex Kruckmnan's example, we do have $$ \forall n\,[(n\text{ is a natural number}\implies n\text{ is even})\lor (n\text{ is a natural number}\implies n\text{ is odd})], $$ but we cannot infer $$ [\forall n\,(n\text{ is a natural number}\implies n\text{ is even})]\lor [\forall n\,(n\text{ is a natural number}\implies n\text{ is even})]. $$

Similarly, in Nathan Smith's example, the trouble arises from the universal quantifier implicit in "necessarily", which means "in all possible situations". Under the stated hypotheses, it is true that "necessarily, [if there is traffic then it is the morning rush hour or if there is traffic then it is the evening rush hour]", but it is not true that "[necessarily, if there is traffic then it is the morning rush hour] or [necessarily, if there is traffic then it is the evening rush hour]".

Similarly, in the first paragraph of Peter Smith's answer, the notion of logical validity includes an implicit universal quantifier "for all truth assignments" and the problem arises from attempting to distribute this quantifier over a disjunction.


My summary of the two interpretations of your question:

Is $(a \Rightarrow (b \vee c)) \Rightarrow ((a \Rightarrow b) \vee (a \Rightarrow c))$ a tautology?


If $a \vdash b \vee c$ then does $ a \vdash b$ or $a \vdash c$?


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    $\begingroup$ In fact, $(a \Rightarrow (b \vee c)) \iff ((a \Rightarrow b) \vee (a \Rightarrow c))$. $\endgroup$ – Alraxite Jul 18 '13 at 2:24

Ok. Just use an example:

If there is traffic it implies it is the morning rush hour or the evening rush hour.

What this statement says it that if there is traffic, it could be either the morning or the evening.... but it's not necessarily either.

So the statements

traffic implies it's the morning or traffic implies it's the evening aren't true.

  • $\begingroup$ That's not correct. If (traffic impies that it is morning or evening) is true, then ( (traffic implies it is morning) or (traffic implies it is evening) ) is also true. $\endgroup$ – Dancrumb Jul 17 '13 at 22:47

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