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.
Best Answer
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.