# Logic, “and” operator between a set of formulas and a formula

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

• If you already know that the formula is inconsistent then why do you need to check it? – fade2black Jul 1 '17 at 18:58
• @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 – 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.
• 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 $S$. – fade2black Jul 1 '17 at 19:26
• 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 – younes zeboudj Jul 1 '17 at 19:48
• $\alpha$ 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. – fade2black Jul 1 '17 at 20:01
• 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 $S$" – younes zeboudj Jul 1 '17 at 20:09