Formally, why does a logical contradiction have probability zero?
We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms,
$$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$
by the definition of the complement. And by the Kolmogorov Axioms, we see that
$$P(\emptyset) = 0.$$
Your axiom list states that
- Additivity: $P(E_1 \cup E_2)=P(E_1)+P(E_2)$, where $E_1$ and $E_2$ are mutually exclusive.
Note that $A \wedge \neg A$ is mutually exlusive with itself.
Hence $$P(A \wedge\neg A)=P\left((A \wedge \neg A)\cup(A \wedge \neg A)\right) = P(A \wedge\neg A) + P(A \wedge\neg A) $$
As booleans, boolean-valued functions, boolean-valued random variables, or any other similar sort of thing, $A \wedge \neg A$ is the same thing as "false", and $A \vee \neg A$ is the same thing as "true".
Applying this, your question reduces to
Why is $P(\text{true}) = 1$ and $P(\text{false}) = 0$?
Answering this depends on the exact details you want things in terms of, but basically $\text{true}$ is always true and $\text{false}$ is never true.
e.g. if, when $X$ is a boolean-valued random variable you define $P(X)$ to mean the probability of the event consisting of exactly those samples where $X$ is true, then $P(\text{true})$ would be the probability of the entire sample space.