# Why is this expression (A and ¬A |= C) entailed?

Hopefully I am posting this in the right place, I am currently in a course of knowledge representation, and I came across an exercise about entailment: $$A\land\neg A\vDash C\,.$$

I would argue that this expression is not entail, but it is actually entailed, but I don't see how, can you help me figure out why?

• This follows from an application of en.wikipedia.org/wiki/Principle_of_explosion Apr 16, 2018 at 15:45
• Using "entailed" in the passive voice without mentioning what entails it seems a bit wrong to me. It seems to me that "I would argue that A∧¬A does not entail C" would be clearer phrasing. Apr 16, 2018 at 21:58

The statement $X\vDash Y$ means "every assignment to the variables that makes $X$ true also makes $Y$ true." Or to put it another way, "There is no assignment of variables that makes $X$ true but fails to make $Y$ true." Well, there's no assignment of variables that makes $A\land\neg A$ true, so there's certianly no assignment that makes $A\land\neg A$ true and also makes $C$ false. So $A\land\neg A\vDash C$ is a true statement.

False entails anything is a rule, analogous to false implies anything.

• Note that the rule "false entails anything" is here used on the metatheory level because $X \models Y$ is defined there, not in the actual logic itself. Even if your (weird) logic without that rule made you have $\bot \rightarrow \bot$ unsatisfiable (i.e. no model exists), this answer would hold. Apr 16, 2018 at 17:19
• @ComFreek Completely agree but I didn't want to go into that level of detail to answer the question "Why does false entail anything?" Apr 16, 2018 at 17:21
• Additionally, you could refer to the Latin proverb "Ex falso (sequitur) quodlibet" - Wikipedia Apr 16, 2018 at 18:24