## lo.logic – Disjunction in weakened Robinson arithmetic

Let $$T$$ denote the theory obtained by removing the axiom $$forall x ( x = 0 lor exists y , S y = x )$$ and restricting double negation elimination to disjunction-free formulas of Robinson arithmetic. In other words, $$T$$ is axiomatized over intuitionistic logic by arithmetical axioms of Robinson arithmetic other than $$forall x ( x = 0 lor exists y , S y = x )$$, and double negation elimination for disjunction-free formulas.

I’ve been playing around with $$T$$ for quite a while (actually months, on and off), and it seems that no genuine theorems containing a positive disjunction can be proven in $$T$$. By genuine I mean those theorems that are not proven by mere logic, like those of the form $$A rightarrow A lor B$$, or $$A lor B rightarrow C lor D$$ where $$A$$ implies $$C$$ and $$B$$ implies $$D$$. For example, I couldn’t find any way of proving $$forall x ( x < 2 rightarrow x = 0 lor x = 1 )$$. As neither induction nor classical logic is at hand, non of the usual ways of thinking seems to work, at least as far as I could check. I’m really not sure about unprovability of this sentence, as I couldn’t find any useful method to show that it’s not provable.

So to avoid any possible complexity that could come from considering more general theorems containing positive disjunctions, I ask my question like this:

Is there a method for showing that $$forall x ( exists y , x + S y = 2 rightarrow x = 0 lor x = 1 )$$ is provable/unprovable in $$T$$?

## Verification of the disjunction between the sub-assemblies of a poset

If there is a poset $$(P, le)$$ and two sets $$X subseteq P$$ and $$Y subseteq P$$, and we have a way $$f: P ^ 2 to 2$$ to efficiently calculate for everything $$(x, y) in P ^ 2$$ if there is a $$z in P$$ such as $$(x le z) wedge (y le z)$$, we want to come back $$mathbf {T}$$ if there is a pair $$(x, y) in X times Y$$ such as $$f (x, y) = 1$$ and $$mathbf {F}$$ otherwise, using as few calls as possible $$f$$ and $$le$$.

## information theory – Upper limit for disjunction together in product distributions

I came across this draft of a manual, where Exercise 6.8 mentions that the disjunction of 2-part ensemble can be solved with a expected number of $$O (n ^ {2/3} log ^ 2 n)$$ bits if Alice and Bob's inputs are sampled independently.

Consider the following protocol. If there is a coordinate $$j in (n)$$ such as $$H (X_j)$$ and $$H (Yj)$$ are both at least $$epsilon$$, then Alice and Bob communicate $$X_j$$,$$Y_j$$. They condition the values ​​they see and repeat this step until no such coordinate can be found. At this point Alice and Bob use Shannon's coding theorem to code $$X$$, $$Y$$. Show how to adjust $$epsilon$$ so that the expected communication can be limited by $$n ^ {2/3} log ^ 2 n$$. Tip: use the fact that $$H (X_j) ge epsilon$$ implies that $$Pr (X_j = 1) ge Omega ( epsilon / ( log (1 / ε)))$$.

I guess the idea is to first communicate all the clues where $$X$$ and $$Y$$ have a large entropy, then use the fact that the remaining clues should have a small entropy. However, the details of the protocol and where the independence of $$X$$ and $$Y$$ is coming, is not clear to me.