<p><span class="math-container">$(x&lt; y) \land (y &lt; z) \land (z &lt; x)$</span></p>

CC BY-SA 4.0
2021-05-09T06:54:36.080Z
2021-05-09T06:54:36.080Z
D.W.
<p>I'm stuck on trying to find an unsatisfiable conjunction of the form <span class="math-container">$a \wedge b \wedge c$</span> where:</p>
<ul>
<li><span class="math-container">$a \wedge b$</span> is satisfiable</li>
<li><span class="math-container">$a \wedge c$</span> is satisfiable</li>
<li><span class="math-container">$b \wedge c$</span> is satisfiable</li>
<li><span class="math-container">$a, b, c$</span> are boolean literals from Linear Integer Arithmetic, i.e. <span class="math-container">$x \leq y$</span>, <span class="math-container">$\neg(3 = 5)$</span>, <span class="math-container">$z = z$</span>, etc.</li>
</ul>
<p>Is there no such case that this is possible (is there a proof for it), or am I just missing an obvious example?</p>

<lo.logic><sat><boolean-formulas>
Linear Integer Arithmetic Satisfiability with Three Literals
gust
# Linear Integer Arithmetic Satisfiability with Three Literals

(Asked by: username_0 on 2021-05-09)

I'm stuck on trying to find an unsatisfiable conjunction of the form $a \wedge b \wedge c$ where:

* $a \wedge b$ is satisfiable
* $a \wedge c$ is satisfiable
* $b \wedge c$ is satisfiable
* $a, b, c$ are boolean literals from Linear Integer Arithmetic, i.e. $x \leq y$, $\neg(3 = 5)$, $z = z$, etc.

Is there no such case that this is possible (is there a proof for it), or am I just missing an obvious example?

**Tags:** lo.logic, sat, boolean-formulas

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## 1 answers

### Answer by username_1 on 2021-05-09. Score: 2 | Accepted answer

$(x< y) \land (y < z) \land (z < x)$


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