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From my understanding these two statements are logically equivalent

p → q ≡∼p ∨ q (can someone 'explainlikei'mfive' why that makes sense)

When I come across this,

(∀a)(∀b)(∃c)[a < b → a < c < b]

is it the same as

(∀a)(∀b)(∃c)[a ≥ b ∨ (a < c < b)]

Temp Jon
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  • I read your question now. $\forall a\forall b\exists c(a<b\Rightarrow a<c<b)\not\equiv \forall a\forall b \exists c(a\ge b\vee a<c<b)$ because you have to negate also quantifiers. – MBYagbasan Nov 03 '17 at 21:52

2 Answers2

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Equivalently and more intuitive is $p\implies q \equiv \neg[p \land \neg q]$

Consider the sentence: If it is raining then it is cloudy. A common misperception is that this means that cloudiness causes rain, or that rain causes cloudiness. Neither is the case. [Edit: It just means that, at a given instant in time, it is not both raining and not cloudy. There is no suggestion of the passage of time or an historical record.]

$Raining \implies Cloudy \equiv \neg [Raining \land \neg Cloudy]$

Dan Christensen
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  • So for the case in my example, could I just substitute one with the other? – Temp Jon Feb 17 '15 at 06:16
  • Yes, whichever is the most convenient form to use in a given situation. If you are at all confused, I usually find it's best to use the conjunctive form. Sometimes I have found that the disjunctive form works well in proofs by induction, e.g. if you are trying to prove an implication by induction. – Dan Christensen Feb 17 '15 at 07:03
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We know that $p \rightarrow q$ is false if and only if $p$ is true and $q$ is false. This can be stated as

$$\neg (p \rightarrow q) \equiv p \land \neg q$$

Negating both sides gives

$$p \rightarrow q \equiv \neg (p \land \neg q)$$

$$\equiv \neg p \lor \neg \neg q \mbox{ (De Morgan's law)}$$

$$\equiv \neg p \lor q$$

$$\therefore p \rightarrow q \equiv \neg p \lor q$$

You can verify this using a truth table.

Muster Mark
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