i have recently seen a question that is stated as " p does not implies q " but due to confusion i have seen a similar question on stack exchange but there is a problem on stack exchange which i do not understood, so here it is .

" p implies q " means if "p" is true then "q" must be true . taking this similar analogy for " p does not q", the statement means if P is true then "q" may or may not be true ( as implication does not hold ) and in both cases the argument will be true unlike for " p implies q" where it becomes false when "P" is true and "q" is false.

Now the answer that i have seen said that " P does not implies q " means this " - ( P -> Q ) " . but normally it means that ( p ->q ) is not always true . which to me does not make any sense because the same could be said for "P implies q " that is also not true always .

here i may have written something wrong but that is what i know . So please tell me the difference between the both , hopefully a truth table will clarify more . And if ambiguity exists somewhere here then please let me know in what circumstances the care should be taken.

  • $\begingroup$ $\neg(p\to q)$ means that $p\to q$ is false. That is, it is true only in the case that $p$ is true and $q$ is false. $\endgroup$ – saulspatz Feb 16 at 18:17
  • $\begingroup$ " p implies q " means if "p" is true then "q" must be true. NO, if you are reading "implies" as the connective "if..., then...". $\endgroup$ – Mauro ALLEGRANZA Feb 16 at 18:29
  • $\begingroup$ "the same could be said for "p implies q " that is also not true always". Exactly; check the truth table for $\to$ : it is not always true. $\endgroup$ – Mauro ALLEGRANZA Feb 16 at 18:45
  • $\begingroup$ @MauroALLEGRANZA i did not get it but yes i am taking implies as "if ... then connective". Further i still did not get the difference between both . Better if you could tell me the logical formula of the both cases. $\endgroup$ – Noob Feb 16 at 18:46
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    $\begingroup$ $(p \to q)$ means : it is false that $p$ is true and $q$ is false. Its negation $\lnot (p \to q)$ will be : it is true that $p$ is true and $q$ is false, i.e. $(p \land \lnot q)$. $\endgroup$ – Mauro ALLEGRANZA Feb 16 at 18:50

The issue you seem to be having is confusing truth with validity. In particular, you are confusing $\neg(P\to Q)$ being true (and thus $P\to Q$ being false) with $P\to Q$ being invalid. (Instead of "valid", it is more likely that you've seen "is a tautology".)

When we're going through a logical argument, the atomic propositions, $P$ and $Q$ in this case, are known or assumed to hold some particular truth values. This is typically formalized by saying we have some truth assignment for all the atomic propositions. We can then calculate what the truth value for a whole formula, e.g. $P\to Q$, is given that truth assignment.

In early logic classes, you are often being asked to show that a given formula is or is not a tautology, i.e. that it is or is not valid. Doing this means showing that the formula is true for all truth assignments. As I said in the first paragraph, you seem to be confusing "$\varphi$ is false" with "$\varphi$ is invalid", probably because you've been given exercises to show that some formula is "false" that should have been stated as, "show that some formula is not a tautology".

So $\neg(P\to Q)$ being valid doesn't mean $P\to Q$ is invalid (which is what "not always true" means); it would mean that $P\to Q$ is always false. Since $P\to Q$ is not always false, neither $P\to Q$ nor $\neg(P\to Q)$ are valid. This situation is sometimes described as $P\to Q$ being contingent. On the other hand, if $\neg(P\to Q)$ is true with respect to some truth assigment, then $P\to Q$ is false with respect to the same truth assignment by definition of $\neg$. Two (propositional) formulas are (semantically) equivalent if they have the same truth value for each truth assignment. Again, by definition of $\neg$, this will never happen with a formula and its negation.

  • $\begingroup$ thanks highly appreciated $\endgroup$ – Noob Feb 17 at 5:22

The Meaning of IMPLIES in Classical Logic

If $p$ and $q$ are logical propositions of unambiguous truth values (either true or false), then

  • $p \implies q$ does not mean that $p$ causes $q$, or that $q$ causes $p$.

  • $p \implies q$ means only that, at the moment, it is false that both $p$ is true and $q$ is false. No other logical connection is assumed.

Example: Let $p$ be the proposition that it is cloudy. Let $q$ be the proposition that it is raining. If, at the moment, $p$ is false (it is not raining) and $q$ is false (it is not cloudy), then we can infer that $p\implies q$, i.e. that if it is raining, then it is cloudy.

Yes, it is counter-intuitive, but this form of logical implication works as the basis for most if not all of modern mathematics. There is no passage of time or cause and effect in mathematics. These are in the realms of science.

Compare the Truth Tables

enter image description here enter image description here Source

Note that $p\implies q$ is false if and only if $p$ is true and $q$ is false (see line 2).

Also, $\neg (p \implies q)$ is true if and only if $p$ is true and $q$ is false (see line 2).

Other interesting points about the truth table for $p \implies q\space$:

  • If $p$ is false then $p\implies q$ regardless of the truth value of $q$ (see lines 3 and 4).
  • If $q$ is true then $p\implies q$ regardless of the truth value of $p$ (see lines 1 and 3).
  • $\begingroup$ thanks for the help $\endgroup$ – Noob Feb 17 at 5:23

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