Script discussed in the video:
https://youtu.be/tqmtZ82WHJc
==== Bayesian calculus vs Propositional logic ====
Idea: Logical implications are akin to Conditional probabilities.
Reading:
$P(B \vert A)=1 \iff A\to B$
$P(B \vert A)=0 \iff \neg(A\to B)$
"$P(B \vert A)$ ... $P(A\to B)$"
Surely
$P(\neg A \lor B) \ge P(B)$
Suggests
$P(A\to B) \ge P(B)$
Bayes' rule gives proportionality factor $\propto P(B)$ above:
$P(B\vert A)\cdot P(A) = P(A\vert B)\cdot P(B)$
So, concretely
$P(B\vert A) = \dfrac{P(A\vert B)}{P(A)}\cdot P(B)$
Or
$P(B\vert A) = \dfrac{P(A\vert B)}{\sum_H^\text{partition} P(A\vert H)\cdot P(H)}\cdot P(B) = \dfrac{P(A\vert B)}{{\mathbb E}_B[P(A\vert B)]}\cdot P(B)$
And also clearly
$P(A\to B) = r_{A,B} \cdot P(B\to A)$
Conditional probabilities as e.g. for probability measures
$P\colon {\mathcal P}T\to[0,1], P(T)=1$
$A,B\subset T$
$P(A)\neq 0\ ...\ P(B \vert A) := \dfrac{P(B\cap A)}{P(A)}$
(Defined only when $A$ isn't a null-set, like $A={}$.)
This also Immediately implies Bayes' theorem.
Note: If $P(B)\neq 0$, then
$P(B \vert A) = \dfrac{P(B\cap A)}{P(A)\cdot P(B)}\cdot P(B)$
Warmup round without the conditional probability notion:
Assuming explosion, we get a propositional version of the Disjunctive Syllogism,
$(\neg A\lor B)\to (A\to B)$
With this
$\neg (A\to B)\leftrightarrow(\neg \neg A\land \neg B)$
In a bivalent truth table (for which, side note, $\neg \neg A = A$), the right hand side,
a conjunction, exactly corresponds to the one failure case of implication.
With the Law of Total probability in some probability theory
$P(Q) + P(\neg Q) = 1$
(Compare with LEM, $Q\lor\neg Q$)
we got
$P(A\to B) = 1 - P(\neg \neg A\land \neg B)$
"There's a chance that $A$ implies $B$, or otherwise we have the one failure case."
Extremal cases:
$\bullet$ $P(T \vert A) = \dfrac{P(T\cap A)}{P(A)} = 1 = P(T)$
("Trivial conclusion is unconditionally true.")
Compare with
$(A\to \top) \leftrightarrow \top$
(But note that we ruled out null $A$, i.e. we assumed $P(A)>0$,
while the propositional formulas, assuming prove the
logical equivalent, here and below, also for false $A$.)
$\bullet$ $P(B \vert T) = \dfrac{P(B\cap T)}{P(T)} = P(B)$
("Trivial assumption doesn't improve credence.")
Also compare with
$(\top \to B) \leftrightarrow B$
Inclusion-exclusion:
$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$
Or
$P(X \cap Y) = P(X) - \delta$ with $\delta = P(X \cup Y)-P(Y)$.
By monotonicity, $\delta\ge 0$, so $P(Y)=1$ kills off the second terms.
$\bullet$ $P(A) = 1\ \to\ P(B \vert A) = P(B)$,
generalizing the above.
But also
$\bullet$ $P(A) = 1\ \to\ P(A \vert B) = 1$
("Credence of something established to be true is
not lowered in light of further assumptions.")
Compare
$A\to (B\to A)$
(Implication introduction)
Particular to the set approach:
$P(B) + P(B^c) = 1$
and further via intersection distributing over unions
$P(B \vert A) + P(B^c \vert A) = 1$
(Law of total probability, independent of conditionals.)
Compare with $B\lor\neg B$, as noted previously,
or also equivalently, the (only formally) stronger
$(A\to B)\lor(A\to \neg B)$
So recall
$P(B), P(A \vert B) = P(A), P(B\vert A)$
(Classical, arithmetic reasoning)
$a,b\in [0,1]$
$\vdash a\cdot b=1\implies a=1\land b=1$
$\vdash a\cdot b=0\implies a=0\lor b=0$
So
$\big(B\land(B\to A)\big) \leftrightarrow \big(A\land(A\to B)\big)$
and classically
$\big(\neg B\lor\neg (B\to A)\big) \leftrightarrow \big(\neg A\lor\neg (A\to B)\big)$
$P(B)=P(A)>0$, then $P(A \vert B) = P(B\vert A)$.
Akin to
$(B\leftrightarrow A)\leftrightarrow\big((B\to A)\leftrightarrow(A\to B)\big)$
(Although this is a general tautology)
Studying negations...
$P(A \vert \neg B)\cdot P(\neg B) = P(\neg B\vert A)\cdot P(A)$.
$P(A \vert \neg B)\cdot \big(1-P(B)\big) = \big(1-P(B\vert A)\big)\cdot P(A)$.
$P(\neg B)>0$:
$P(A\vert \neg B) = P(A)\cdot\frac{1-P(B\vert A)}{1 - P(B)}$
$P(\neg A\vert \neg B) = 1 - P(A)\cdot\frac{1-P(B\vert A)}{1 - P(B)}$
$P(A)>0$:
$P(B\vert A) = 1\ \leftrightarrow\ P(\neg A\vert \neg B) = 1$
So (under the constraint that we didn't consider the explosive antecedents)
$(A \to B) \leftrightarrow (\neg B \to \neg A)$.
