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

  • $\begingroup$ 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

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).

  • $\begingroup$ 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

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