A note on the law of identity and the converse Parry property
Proof: 1.1. Every mef will be of the form p—> p in virtue of variablesharing. Eefs of the form p —> C and C -+ p, where C is an ef, are ruled out, the first because of the Ackermann property, the second because by modus ponens p would be a theorem. So every eef is a mef, and Lemma 1.1. follows. 1.2. Let A contain two or more propositional variables. In virtue of variable-sharing every ef of A containing two or more propositional variables will have on at least one side of its arrow a subformula containing at least two propositional variables (so this subformula will be an ef). (E.g., imagine an ef B containing two propositional variables, p and q; then it will contain p either on both sides of the arrow, in which case q is on at least one side, or on only one side, in which case q must be on both.)*