Can you calculate the probability of a complex logic expression?

[yeet991only]

TL;DR

Any logical expression can be simplified to AND , NOT , OR. But these are same as intersection , negation and reunion in probability.

Suppose this expression : ((A ∨ ~B) ∨ (C -> D))
This simplifies to : A + ~B + ~C + D , using boolean algebra.
Now suppose i know that this expression is true: p( ((A ∨ ~B) ∨ (C -> D)) ) = 1;
and I also know that p(A) = 0.3 , p(B) = 0.99 , p(C) = 0.92 .
but the probability of the expression is actually: p(A ∪ ~B ∪~C ∪ D) = 1
So can you calculate D from that? (Normally you could, but the question is if combining logic and probability like this is valid).

Also clearly (C -> D) isnt the kind of arrow that you would use in a bayesian network or stuff, because it reduces to p( ~C or D) , but in graphs it is the bayes formula for such arrow : p(C | D) = p(D|C) * p(C) / p(D).

 

[Dale]

TL;DR Summary: Any logical expression can be simplified to AND , NOT , OR. But these are same as intersection , negation and reunion in probability.

So can you calculate D from that?

No. From just the information given we don’t know if the probabilities of one are dependent on the others. Without knowing that they are independent or mutually exclusive or knowing their conditional probabilities, $p(D)$ cannot be calculated from the given information

 

[jbergman]

TL;DR Summary: Any logical expression can be simplified to AND , NOT , OR. But these are same as intersection , negation and reunion in probability.

Suppose this expression : ((A ∨ ~B) ∨ (C -> D))
This simplifies to : A + ~B + ~C + D , using boolean algebra.
Now suppose i know that this expression is true: p( ((A ∨ ~B) ∨ (C -> D)) ) = 1;
and I also know that p(A) = 0.3 , p(B) = 0.99 , p(C) = 0.92 .
but the probability of the expression is actually: p(A ∪ ~B ∪~C ∪ D) = 1
So can you calculate D from that? (Normally you could, but the question is if combining logic and probability like this is valid).

Also clearly (C -> D) isnt the kind of arrow that you would use in a bayesian network or stuff, because it reduces to p( ~C or D) , but in graphs it is the bayes formula for such arrow : p(C | D) = p(D|C) * p(C) / p(D).

A related area is probabilistic logic programming https://en.m.wikipedia.org/wiki/Probabilistic_logic_programming

 

[D H]

TL;DR Summary: Any logical expression can be simplified to AND , NOT , OR. But these are same as intersection , negation and reunion in probability.

Any logical expression can also be simplified to NAND only, NOR only, and if one allows $c\to\text{false}$, also by implication only as $c\to\text{false}$ is NOT c.

Suppose this expression : ((A ∨ ~B) ∨ (C -> D))

That expression can be write as $((A\vee \neg B) \vee (\neg C \vee D))$, and now one can dispense with the parentheses, yielding $(A\vee \neg B \vee \neg C \vee D)$ -- and also $((A \vee \neg B \vee \neg C) \vee D)$. I'll use that final form in just a bit.

Now suppose i know that this expression is true: p( ((A ∨ ~B) ∨ (C -> D)) ) = 1;
and I also know that p(A) = 0.3 , p(B) = 0.99 , p(C) = 0.92 .
but the probability of the expression is actually: p(A ∪ ~B ∪~C ∪ D) = 1
So can you calculate D from that? (Normally you could, but the question is if combining logic and probability like this is valid).

One obvious solution is that D is always true, in which case P(D) = 1. There are other solutions. Note that it doesn't matter what D is in the case that $(A \vee \neg B \vee \neg C)$ is true. However, we must have D being true whenever $(A \vee \neg B \vee \neg C)$ is false. Note that D always being true satisfies this requirement, but so does $D = \neg(A \vee \neg B \vee \neg C) = (\neg A \wedge B \wedge C)$. Given the sited probabilities, that D is true (but not necessarily always true) seems quite likely.