Timeline for Why is A implies B true if A is false and B is false?
Current License: CC BY-SA 3.0
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| Nov 27, 2016 at 23:48 | comment | added | durron597 | @DavidRicherby let $r = \neg q$. Clearly $(p \wedge \neg p) \rightarrow q$ is just as valid as $(p \wedge \neg p) \rightarrow r$. From falsehood, anything follows, including another contradiction. | |
| Nov 27, 2016 at 13:28 | comment | added | David Richerby | I don't buy this argument. You're saying that, by writing $P\to Q$ where we know $P$ to be false, we're talking about some alternate reality where $Q$ could be true. If that's the case, why do you then go on to assume that $Q$ is true in that alternative reality? That seems philosophically unsatisfactory. Also, the whole "alternative reality" setup completely contradicts the formal semantics of logics: the truth or falsity of a formula in a particular model is determined with respect to that model, not with respect to some other model that the reader dreams up. | |
| Nov 27, 2016 at 10:52 | review | First posts | |||
| Nov 27, 2016 at 13:28 | |||||
| Nov 27, 2016 at 10:49 | history | answered | durron597 | CC BY-SA 3.0 |