In the preceding section (with oxford scholarship), Field has outlined a generalisation of Łukasiewicz’s continuum valued semantics. And similar to this, it provides a operator `D’ (read `determinately’) with the following properties:
ia) v(A)=1 then v(DA)=1
ib) v(A) = 0 then v(DA)=0
ic) 0 < v(A) < 1 then v(DA) < v(A)
ii) v(A) 
Field adds a stronger alternative to (ib):
(ib-s) v(A) 
Field points out that these semantical properties corresponds on one hand to the algebraical properties of an operator \delta on the value space. On the other hand, they represent the following meta-logical statements, which he calls natural laws of inference. Field only sketches the respective derivations, I will be a bit more explicit. Aaccording to his convention from §10.6 I read the turnstile as weak validity, that is, for preservation of value 1.
(1) |= DA -> A
This is a direct consequence of (ib) which excludes `DA’ to have value 1 when A does not.
(2) A |= DA
(ia) guarantees that whenever v(A)=1, v(DA)=1, too.
(3) if |= A -> DA then |= A v ¬A
Assume that v(A -> DA)=1 and v(A v ¬A) <1. The latter means that sup{v(A), v(A)*}<1, hence v(A)<1. Assuming the Łukasiewicz semantics for the conditional this excludes v(A -> DA) being 1, contradiction. Therefore v(A v ¬A) = 1.
(4) if |= A -> ¬A then |= ¬DA
For v(A->¬A) to be 1 v(A) must be less than or equal to v(¬A). Following (ib-s) this renders v(DA)=0 and v(¬DA) accordingly 1.
(5) If |= A->B then |= DA -> DB
If v(A->B)=1 then v(A)
Although Field has not been explicit about this in this chapter so far
(3) (4) and (5) show that he connects with the definition of the conditional.
To provide the conditional with these properties Field defines DA as A & ¬(A -> ¬A). That `D’ meets the above requirements (ia) – (ii) thus would ensured by the necessary conditions for a conditional (I – III). Again, I will try to spell out the arguments to which Field merely suggests.
(ia)
assume that v(A)=1. According to the semantics, this makes v(A)*=v(¬A)=0. With III this amounts to v(A -> ¬A)=0. Again, v(A -> ¬A)*= v(¬(A -> ¬A)) = 1. Because inf{1, 1}=1, it therefore holds that inf{v(A),v(¬(A -> ¬A))} = 1. Hence, v(A)=1 then v(A & ¬(A -> ¬A))= v(DA) = 1.
(ib)
assume that v(A)=0. Then v(DA)=v(A & ¬(A -> ¬A)) = inf{0, v(¬(A -> ¬A))}=0.
(icw)
assume that 0 < v(A) < 1. v(DA)=v(A & ¬(A -> ¬A)) = inf{v(A), v(¬(A -> ¬A))}. Now, either v(DA)=v(A), in which case v(DA) \leq v(A). Or v(DA) = v(¬(A -> ¬A)), but only if v(¬(A -> ¬A)) < v(A), why in this case v(DA) < v(A) and therefore v(DA)
(ib-s)
Assume that v(A) \leq v(¬A). With (I) this is equivalent to that v(A -> ¬A)=1, and accordingly v(¬(A -> ¬A)) = 0. Hence inf{v(A),v(¬(A -> ¬A))} = v(DA) = 0.
(ii)
With this result I struggle, although Field doesn’t consider it especially. Given that v(A)
| v(A) | v(¬(A -> ¬A)) | |
| v(B) | (1) | (2) |
| v(¬(B -> ¬B)) | (3) | (4) |
Case (1) is straightforward, if v(DA) = v(A) and v(DB) = v(B) then v(DA)
Case (2) is equally simple, given that it presupposes v(¬(A -> ¬A)) < v(A).
For (3) and (4), however, I don’t see the solution. Field remarks that IIa and IIb from above would guarantee (ii). I assume that if anywhere, they need to be deployed in these third and fourth case. Let’s focus on (3). (IIb) gives, e.g., v(B -> ¬A)
So I’m stuck at this point. Can anybody help me solve this such that I can continue and proceed to the actually relevant, and actually difficult parts?
