1
$\begingroup$

Consider a set $S$ of formulas $\beta_i$ and a formula $\alpha$, if we have a condition such as $S \land \alpha$ is inconsistent what we have to calculate to check the inconsistency of $S \land \alpha$? In other words, what is the result of $S \land \alpha$?

For example, $S=\{\beta_1, \beta_2, \beta_3\}$

|cite|improve this question
$\endgroup$
  • $\begingroup$ If you already know that the formula is inconsistent then why do you need to check it? $\endgroup$ – fade2black Jul 1 '17 at 18:58
  • $\begingroup$ @fade2black i don't know if the formula is inconsistent, it is a condition, something like $if( (S\land \alpha) is consistent) then ...$ so i have to check the inconsistency of the result $\endgroup$ – younes zeboudj Jul 1 '17 at 19:04
1
$\begingroup$

A set $S$ of formulas (propositional logic) is consistent if there is a truth assignment under which all formulas in $S$ are true, in other words if there is a model of $S$. Alternatively, $\{\beta_1, \beta_2, ..., \beta_n \}$ is consistent, iff $\beta_1 \wedge \beta_2 \wedge ... \wedge \beta_n$ is satisfiable.

You could just find a truth assignment such that $\alpha \wedge \beta_1 \wedge \beta_2 \wedge \beta_3$ is true. If the number of variables is small then you can check it by hand by creating a truth-table, or use Method of analytic tableaux. Also, note that the Satisfiability problem (SAT) is NP complete.

You can also use a resolution procedure to show that a formula is unsatisfiable (inconsistent).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ So having one beta of S inconsistent with alpha make S and alpha inconsistent regardless of other betas $\endgroup$ – younes zeboudj Jul 1 '17 at 19:09
  • $\begingroup$ Right. In addition, when we say that $S=\{\beta_1,...,\beta_n\}$ is consistent we mean that $\beta_1 \wedge... \wedge\beta_n$ is satisfiable. But if $\alpha$ is not consistent with $S$ it does not yet mean that $\alpha$ is inconsistent with a subset of $S$. $\endgroup$ – fade2black Jul 1 '17 at 19:26
  • $\begingroup$ How is that!, if $\alpha$ is not consistent with $S$ then forcly $\alpha$ is inconsistent with a subset of $S$ except when $\alpha$ is inconsistent $\endgroup$ – younes zeboudj Jul 1 '17 at 19:48
  • $\begingroup$ $\alpha$ is consistent with $S$ means $\alpha \wedge \beta_1 \wedge \beta_2 \wedge \beta_3$ is satisfiable. So if $\alpha$ is inconsistent, say, with $\beta_2$ then $\alpha \wedge \beta_2$ is not satisfiable (always false), and hence $\alpha \wedge \beta_1 \wedge \beta_2 \wedge \beta_3$ is false too (independent of $\beta_1$ and $\beta_3$). So having one $\beta$ inconsistent with $\alpha$ makes $\alpha$ inconsistent with $S$. However, $S' = \{\beta_1, \beta_3\}$ may be consistent with $\alpha$ as long as $\alpha \wedge \beta_1 \wedge \beta_3$ is satisfiable. $\endgroup$ – fade2black Jul 1 '17 at 20:01
  • $\begingroup$ I'm okay with your last comment, but please review your first one, you said "... But if $\alpha$ is not consistent with $S$ it does not yet mean that $\alpha$ is inconsistent with a subset of $S$" $\endgroup$ – younes zeboudj Jul 1 '17 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.