[FOM] Weak logic axioms

Michael Lee Finney michael.finney at metachaos.net
Thu Sep 24 21:32:26 EDT 2015


 I think that truth does distribute. I am assuming that T(p) means that
 p is "distinguished" and therefore true.

 T(p & q -> r) holds exactly when one of the following conditions
 occur.

    (1) T(p) does not hold
    (2) T(q) does not hold
    (3) All of T(p), T(q) and T(r) hold

 So if (1) is the case, then T(p -> r) holds and so T(p -> r) v T(q -> r)
 holds regardless of T(q). Symmetrically, the same is true if (2)
 is the case.

 If (3) is the case then both T(p -> r) and T(q -> r) trivially hold.

 So, as far as I can see, T(p & q -> r) -> T(p -> r) v T(q -> r)
 and so distribution of truth over disjunction is not invalidated.

 On another note, your method of reading (p & q -> r) -> (p -> r) v (q -> r)
 does not work when contraposition is not valid.

 Since my original post, I have learned more about these theses. They
 all derive from the axiom

   (p -> q) v (q -> p)

 known as the "chain", "pre-order" or "linear" axiom. This is not a
 very popular axiom, but what it is really doing is constraining what
 happens to the implication (p -> q) when p is not distinguished. It
 says that if p is not distinguished then (p -> q) is distinguished.

 If both p and q are distinguished, then trivially both (p -> q) and
 (q -> p) are distinguished. The question is what happens when either
 p or q (or both) is not distinguished. If p is not distinguished
 there are two cases (symmetrically for q, of course).

   A) q is distinguished, so (q -> p) cannot be distinguished and
      therefore (p -> q) must be distinguished.

   B) q is not distinguished, so both (p -> q) and (q -> p) must be
      distinguished.

 Other axioms don't address this semantic property of entailment. They
 only address what happens when values are distinguished.

 This assumes primality because if the logic is not prime then we
 could have

    T((p -> q) v (q -> p))

 but with neither T(p -> q) nor T(q -> p) holding.

 It turns out that this axiom is the "root" of all conjunctive and
 disjunctive distribution principles (the hard directions, at least).
 In fact, the derivations go from

   (p -> q) v (q -> p)

 to

   (p -> q v r) -> (p -> q) v (p -> r)

 to

   p & (q v r) -> (p & q) v (p & r)

which makes it very interesting in a non-relevant weak logic.



>  
>         Some time ago under the present subject heading Michael Lee Finney wrote:
> "...you could then prove
>  (4)   (p & q -> r) -> (p -> r) v (q -> r)
> which I thought that surely was invalid. But again, I appear to
> have been incorrect. I asked if these are classical theorems, and it
> turns out that they are amazingly easy to prove in classical logic
> (just turn entailment into disjunction and you are pretty much there)."
>          I would like to further comment on (4): 
> There seems to be however no understandable way of reading (4)
> which makes it true. Bur if the disjunctions in the consequence are transposed, thus:
>  '(-r->-p) v(-r->-q)' then (4) is understandable. The consequence
> may mistakenly be read as '(pvq)->r'. One may question (4) on the
> ground that that the antecedent does not imply either disjunct. That
> would be a mistake as well for we have the same with 
'p->>(qv-q)'. 
>         (4) is interesting for it indicates that Truth does not
> distribute over disjunction. For if 'T[p & q) -> r) -> (p -> r) v (q
> -> r)]' then 'T(p & q -> r) -> T[(p -> r) v (q -> r)]'. But the following is not true:
(4*)'T(p & q ->> r) -> [T(p -> r) v T(q -> r)]'.

> Alex Blum
>   







>   
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