↧
Answer by Farmer S for Does "agreement on cardinalities" imply second-order...
No, second order logic does not have the weak test property: let $\mathfrak{B}=(\mathbb{R},{<})$ (that is, the real numbers with the only predicate being the usual "less than" order) and let...
View ArticleDoes "agreement on cardinalities" imply second-order elementary...
Say that a logic $\mathcal{L}$ satisfies the weak test property iff for all $\mathfrak{A}\subseteq\mathfrak{B}$ we have $(1)\implies(2)$ below:For each $\mathcal{L}$-formula $\varphi$ with parameters...
View Article