A set of propositional formulas $P_1, \cdots, P_k$ is consistent iff there is an environment in which they are all true.

Write a formula, $S$, so that the set $P_1, \cdots, P_k$ is not consistent iff $S$ is valid.

Sorry for asking this homework question. I really don't know how to solve it.


Let $S := \lnot (P_1 \land \ldots \land P_k)$.

If $\{ P_1, \ldots, P_k \}$ is inconsistent (i.e. unsatisfiable), then $(P_1 \land \ldots \land P_k)$ is always false, and thus its negation is a tautology (i.e. valid).

If $S$ is valid, then $(P_1 \land \ldots \land P_k)$ is always false (a contradiction) and thus there is no truth assignment that can simultaneously satisfy all the $P_i$'s, i.e. $\{ P_1, \ldots, P_k \}$ is unsatisfiable (i.e. inconsistent).

According to your post

a set $F$ of formulas is consistent iff there is a truth-assignment such that all the formulas of $F$ are true.

Thus, inconsistent (the negation of consistent) means that there is no truth-assignment for which the formulas are all true.

This in turn means that there is no truth-assignment that simultaneously satisfies all formulas, and this amounts to saying that the set of formulas is unsatisfiable.

  • $\begingroup$ The proof is pretty good and clear. But I don't understand what the '(i.e. unsatisfiable)' means? $\{P_1, \cdots, P_k\}$ is a set. A set does not poess the property of satisfiability which is the quality of a proposition formula. $\endgroup$ – 王文军 or Wenjun Wang Mar 14 at 6:41

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