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On page 66, A Mathematical Introduction to Logic, Herbert B. Enderton(2ed),

For each of the following conditions, give an example of an unsatisfiable set $\Gamma$of formulas that meets the condition.

(a) Each member of $\Gamma$ is—by itself—satisfiable.

(b) For any two members $γ_1$ and $\gamma_2$ of $\Gamma$, the set $\{γ_1, γ_2\}$ is satisfiable.

(c) For any three members $γ_1$, $γ_2$, and $γ_3$ of $\Gamma$, the set $\{γ_1, γ_2, γ_3\}$ is satisfiable.

Except for (a), I have difficulty in constructing examples of (b) and (c).

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2 Answers 2

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Why not something like: $a\implies b$, $\lnot a\implies c$, $b\implies \lnot a$ and $c\implies a$. You get a contradiction if all $4$ are true, but any three do not lead to contadictions.

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Thank you for your answer. I thought $\Gamma$ should be infinite. –  Metta World Peace Jan 5 '13 at 23:28
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For two, consider the sentences $$A\lor B, \quad A\lor \lnot B, \quad \lnot A\lor B,\quad \lnot A \lor B.$$

Any two are satisfiable, but the collection of all $4$ cannot be. To check satisfiability, it is enough to use the first two, since there is more symmetry than is apparent.

For three, consider all disjunctions of the form $X\lor Y\lor Z$, where $X$, $Y$, and $Z$ respectively range over the atoms $A$, $B$, and $C$, and their negations.

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