Are all substitution-closed sets schematic?

Question

I assume everyone is familiar with propositional logic and its language.

Let $S$ be a set of wffs of propositional logic.

The set $S$ is said to be substitution-closed iff whenever $x$ is in $S$, then $f(x)$ is in $S$, where $f$ is a substitution. A set is said to be schematic iff it is the set of all substitutions of a formula.

These are definitions I coined myself. I have proven that every schematic set is a non-empty substitution-closed set. When I try to prove the converse, I get stuck. I am starting to think there is a counterexample. Can anyone confirm or deny my suspicions?

Sorry, I meant f. I edited it now. – user107952 Dec 24 '13 at 16:12 — user107952, Dec 24 '13 at 16:12

score 0 · Accepted Answer · answered Dec 25 '13 at 01:58

0

Nevermind, I got the answer now. The set of conjunctions union the set of disjunctions is a substitution closed set, but is not schematic.

answered Dec 25 '13 at 01:58

user107952

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Are all substitution-closed sets schematic?

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