In order to understand how to get from D strategies to E strategies, I've been trying to think about cases where they diverge -- that is, dialogues which are D-strategies, but not already additionally E-strategies. Recall that we are in the presence of D11 and D13; by D11, an attack may be defended at most once, and by D13, a P-assertion can be attacked at most once. These put significant constraints on O. In fact, I've come to the conclusion that the only time when O will be faced with a choice is when P has attacked a conditional; then, and only then, O has a choice between defending the conditional and attacking its antecedent (assuming it is non-atomic). In either of these cases, it follows that the conditional was originally asserted by O.

Under what conditions will a conditional be asserted by O? In defense of \wedge_L, \wedge_R, or ?, or as an attack on a negation. The former imply that O has previously asserted a conjunction or a disjunction one of whose conjuncts/disjuncts is a conditional. Take the latter; this means that P has asserted a negated conditional. I have yet to come up with an example where P actually does so, but I'm wondering if it shouldn't be possible to construct one via substitution of such for an atom in a IL-valid formula. Hmmmm...

Well, I've found a formula which should be IL-valid and allows O to assert a conditional:

(implies (not (not (implies (and p p) q))) (not (not (not (not (implies (and p p) q))))))

Or, in more readable format:

(¬¬((P ∧ P) → Q) → ¬¬¬¬((P ∧ P) → Q))

However, at the point where O has a choice of moves, if he *doesn't* take the option of defending the attack on the conditional, then he loses it; the rest of the dialogue is so full of attacks on negations, which are undefensible and hence always remain open, that D11 will prevent him from going back and re-defending. So, in order for a D strategy to differ from an E strategy, (a) O must assert a conditional, (b) P must attack it, (c) O must defend it immediately.

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