**Reasoning from Assumptions Attempt #1 (Naive)**: Fix a set of formulas Γ. Γ\vDashφ iff P has a winning strategy for φ when he is allowed to assert any ψ\inΓ, either as an attack or as a defense. I quickly realized that this was not acceptable, since it doesn't work to capture reasoning from implication in CL; take the consequence (p&q)\vDash(p&q): If P asserts the right-hand side, then the material on the left won't help him at all when O attacks (p&q). This lead to:

**Reasoning from Assumptions (Naive) Attempt #2**: Γ\vDashφ iff P has a winning strategy for φ when he is allowed to attack any &psi&\inΓ and to defend with anything in Γ. Now, if we think that any formula is a consequence of its own assumption in N (which may or may not be a reasonable thought), then this approach doesn't work since it fails to make (pVq)\vDash(pVq); if P asserts pVq (the right-hand side), and O attacks it, then if P counterattacks by attacking pVq (the left-hand side), this allows O to defend, and the dialogue will never end. This is only a problem if we think that this consequence should be valid in N. Assuming we do, let's get a bit more creative:

**Reasoning from Assumptions Attempt #3**: Let a(Γ) be the set of atoms which are subformulas of elements of Γ. Then Γ\vDashφ iff P has a winning strategy for φ when he is allowed to assert anything in a(Γ) whenever he wants. This fails because then (pVq)\vDash(p&q). We should also consider:

**Reasoning from Assumptions #4 (Jesse's proposal)**: Γ\vDashφ iff there is Γ'\subseteqΓ and \vDash\bigwedge Γ'\rightarrow φ. I dislike this suggestion because it builds in the validity of the deduction theorem, which begs the question against the possibility of reasoning from assumptions in N. I haven't come up with any other candidates yet.

Here's the kicker: Any ruleset where P has a winning strategy for (p&p)->p, assumptions need to be handled in such a way that when O concedes p&p, however that gets cashed out, P is able to assert an atomic formula as his initial statement. Otherwise, the deduction theorem will break.

ReplyDelete