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In classical logic, why is $(p \Rightarrow q) = T$ if $p = F$ and $q = F$?
Is the only way to prove that is not a tautology by having $T \to F$, which gives $F$?
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oups i did wrong – ek.Sek Jun 21 '18 at 19:34
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i edited my question – ek.Sek Jun 21 '18 at 19:34
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Yes, the only way $p\implies q$ can be false is if $p$ is true and $q$ is false; phrased differently, the sentence $$\neg(p\implies q)\iff \neg p\wedge q$$ is a tautology. (Here "$\wedge$" means "and" and "$\neg$" means "not," and they each bind more tightly than $\iff$.) Is there anything in the answers to the previously-linked question which is unclear? If so, what? (If not, then what remains unclear? Is there an issue with any of the remaining three options?) – Noah Schweber Jun 21 '18 at 19:40
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@NoahSchweber thank you a lot,you are clear.I understand it,cause the previous is telling a lot and it is confusing. A line like this is understandable.I thought for example False->False it may be produce False as result,but i see it on the truth table that False ->False gives True and i asked my self.I get influenced by F^F=F – ek.Sek Jun 21 '18 at 19:59
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$p{\implies}q$ promises that $q$ will be true if $p$ is. This promise is only broken (falsified) when $q$ is false when $p$ is true. When $p$ and $q$ are both false we can not say the promise is not kept.
"$q$ actually is false, but I may still promise it would be true if $p$ were, although it happens to be that $p$ is not."
PS: We do not say $p{\implies}q$ is a tautology when $p$ and $q$ are false; we just say it is true. A tautology is true for all possible evaluations for the literals.
However, $\bot{\implies}\bot$ is a tautology, since the only literals are the falsity constant whose only permissable evaluation is false.
Graham Kemp
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