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This is the Exercise 7 in 《Logic for Computer Scientist》(Uwe Schoning) 

Exercise 7: Give an example of a 3-element set M so that M is not
satisfiable, but every 2-element subset of M is satisfiable. Generalize your
example to n-element sets.

I am stuck in this question for a while. I intuitively thought it was easy, while it was not for me at least :(.

I think there may be something wrong in my understanding and this answer should help me a lot .

Thanks.

2 Answers 2

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For $3$ elements: $$a,b,(\text{not }(a \text{ and } b))$$ For $n$ elements: $$a_1,a_2,\ldots,a_{n-1},(\text{not }(a_1 \text{ and } a_2 \text{ and }\ldots\text{ and }a_{n-1}))$$

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    the Aha moment occurs i think . Thanks a lot ! – 2017-02-27
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    Another example is $(a \leftrightarrow b) \wedge (b \leftrightarrow c) \wedge (c \leftrightarrow \neg a)$, with its obvious generalization. – 2017-02-27
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    @FabioSomenzi, does it meet the "3 elements in M" ? What is the actual elements ? – 2017-02-27
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    Well, write it like this: $\{a \leftrightarrow b, b \leftrightarrow c, c \leftrightarrow \neg a\}$. Sorry, I'm used to switch between the two forms without even thinking. – 2017-02-27
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    My version of that was: force the simultaneous satisfiability of the first $(n-1)$ elements (hence why not make them all atoms), then worry about the last one. – 2017-02-27
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    @Keith quasi's example is more star-shaped, and mine looks more circular, but both have a rotational symmetry of sorts. So, yes, that's the easy way to get examples that easily generalize to $n$ elements. – 2017-02-27
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    @Keith Here's another example, already in the $4$-element version: $\{a, a \rightarrow b, b \rightarrow c, \neg c\}$. This may be close to what you were looking for. – 2017-02-27
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    Thanks @FabioSomenzi, my first guess about the answer is something "symmetric rotational " form, like $AB\neg{C},A\neg{B}C,\neg{A}BC$. Your answer and quasi's gave me those "simple" forms. I will re-think about this question. I know where i was going wrong now. – 2017-02-27
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    @quasi, would you please help review my answer ? – 2017-03-01
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    @FabioSomenzi, would you please help review my answer ? – 2017-03-01
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After a few hours thinking regarding the answers of quasi & Fabio, here is my answer.

I am willing to find a "pattern" or "constraint" so that any set of formulas meeting this "pattern" will apply to the question.

My induction :

  1. from the first premise " 3 element is not satisfiable", so there are the constraints : $(A\wedge{B}\rightarrow\neg{C}) \wedge (A\wedge{C}\rightarrow\neg{B}) \wedge (B\wedge{C}\rightarrow\neg{A}) $ . After equivalence, it shows the same thing : $A\wedge{B}\rightarrow\neg{C}$.

  2. from the second premise " any 2 elements subset is satisfiable", so we have the constraints: $A\rightarrow{B}, B\rightarrow{C},C\rightarrow{A},B\rightarrow{A},C\rightarrow{B},C\rightarrow{A} i.e. (C\leftrightarrow{B})\vee(A\leftrightarrow{B})\vee(A\leftrightarrow{C})$

  3. join the constraints from step1/step2, we finally get the constraint $A\wedge{B}\rightarrow\neg{C}$.

So for any formula set which met this constraint will apply to the question.

By testing the samples given by quasi&Fabio, it is valid solutions.

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    If $A$, $B$, and $C$ range over formulae, then the mutual exclusion requirement is indeed that $\neg A \vee \neg B \vee \neg C$ should be valid. The second requirement, though, is that $A \wedge B$, $B \wedge C$, and $A \wedge C$ should be (individually) satisfiable. – 2017-03-01
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    @Keith: To check it, just make a truth table with 8 rows, one row for each possible valuation of the triples A,B,C. Of course you need need columns for A,B,C, and for your 3 formulas, as well as the 3 paired combinations and the one triple combo. – 2017-03-01
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    @Keith: It's a little tedious, but then you'll know for sure. – 2017-03-01
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    @quasi, yes, that is what i am using in my draft paper for induction. – 2017-03-01