User:C7X/Example of satisfaction

Since each formula is just a string, we need a way to interpret them semantically for them to be meaningful. This is done by the "satisfaction relation" \(\vDash\). For simplicity we'll do this with a sentence, i.e. a formula with no free variables. (Reason)

Define addition for natural numbers as usual (with \(S(x)=x+1\)). Then, we can demonstrate \(\{1,3,7\}\not\vDash``\exists a\exists b\exists c(a+b=c)"\).

There are only 27 possible variable assignments. We can rule out the assignments for which \(s(``a")>s(``c")\lor s(``b")>s(``c")\), which leaves 14 to be manually checked: So there does not exist a variable assignment \(s\) such that \(\{1,3,7\},s\,\vDash\,``\exists a\exists b\exists c(a+b=c)"\) is true.
 * \(s(``a")=1,\;s(``b")=1,\;s(``c")=1\).
 * \(s(``a")=1,\;s(``b")=1,\;s(``c")=3\).
 * \(s(``a")=1,\;s(``b")=1,\;s(``c")=7\).
 * \(s(``a")=1,\;s(``b")=3,\;s(``c")=3\).
 * \(s(``a")=1,\;s(``b")=3,\;s(``c")=7\).
 * \(s(``a")=1,\;s(``b")=7,\;s(``c")=7\).
 * \(s(``a")=3,\;s(``b")=1,\;s(``c")=3\).
 * \(s(``a")=3,\;s(``b")=1,\;s(``c")=7\).
 * \(s(``a")=3,\;s(``b")=3,\;s(``c")=3\).
 * \(s(``a")=3,\;s(``b")=3,\;s(``c")=7\).
 * \(s(``a")=3,\;s(``b")=7,\;s(``c")=7\).
 * \(s(``a")=7,\;s(``b")=1,\;s(``c")=7\).
 * \(s(``a")=7,\;s(``b")=3,\;s(``c")=7\).
 * \(s(``a")=7,\;s(``b")=7,\;s(``c")=7\).