If $P$ a probability of a sentence to be true, then $\{P(\phi | T_i)\}_{i \in \mathbb{N}}$ is a martingale over constructed theories $T_i$ I am reading Section 2.1 of Definability of Truth in Probabilistic Logic.
For a language $L$, fix a probability distribution $P:L \to [0,1]$. Enumerate sentences $\phi_1, \phi_2, \ldots$ of a language $L$, let the first $L$-theory be $T_0:=\emptyset$. Let $\phi_j$ be the first sentence independent of a theory $T_i$. Then let $T_{i+1}:=T_i \cup \{\phi_j\}$ with the probability $P(\phi_j\mid T_i)$ and $T_{i+1}:=T_i \cup \{\neg \phi_j\}$ with the probability $P(\neg \phi_j\mid T_i)$ where $P(\phi_j\mid T_i) = P(\phi_j \land T_i) / P(T_i)$.
Further suppose that the following holds
$$P(\phi \mid  T_i) = P(\phi \mid  T_i \land \phi_j)P(\phi_j \mid  T_i)+P(\phi \mid  T_i \land \neg \phi_j)P(\neg \phi_j \mid  T_i).$$
Show that the sequence $\{P(\phi | T_i)\}_{i \in \mathbb{N}}$ is a martingale.
My problem is that I do not know probability theory and when looking at the definition of a martingale at Wikipedia, it implies that if a sequence $\{X_i\}_{i \in \mathbb{N}}$ of random variables is a discrete-time martingale, then $E(X_1 | X_0)=X_0$. But I do not even know how to show that $E(P(\phi \mid \phi_j) \mid P(\phi))=P(\phi)$.
 A: Disclaim : I'm not a specialist of probability; thus, this is only an HINT.
See John Walsh, Notes on Elementary Martingale Theory :

Conditional probability of $B$ given $A$ : $P(B | A) = P(A \cap B) /P(A)$
Conditional expectation of $X$ given $A$ : If $B$ is an event then, $I_B$ is a random variable with $P(I_B = 1 | A) = P(B | A)$, and $P(I_B = 0 | A) = 1 − P(B | A)$ so that :

$P(B | A) = E(I_B | A)$.

Definition 2.1 A filtration on the probability space $(\Omega, \mathcal F, P)$ is a sequence $\{ \mathcal F_n : n = 0, 1, 2, \ldots \}$ of sub-sigma fields of $\mathcal F$ such that for all $n, \mathcal F_n \subset \mathcal F_{n+1}$.

In our case, the filtration is the sequence of $\mathcal L$-theories $T_i$ : it is true that $T_i \subset T_{i+1}$.

Definition 2.2 A stochastic process is a collection of random variables defined on the same probability space.

In our case, the formula $\varphi$ is the stochastic process.

Definition 2.3 A stochastic process $X = \{ X_n, n = 0, 1, 2,\ldots \}$, is adapted to the filtration $(\mathcal F_n)$ if for all $n, X_n$ is $\mathcal F_n$-measurable.
Definition 2.4 A process $X = \{ X_n, \mathcal F_n, n = 0, 1, 2,\ldots \}$, is a martingale if for each $n = 0, 1, 2, \ldots$,
(i) $\{ \mathcal F_n, n = 0, 1, 2 \ldots \}$ is a filtration and $X$ is adapted to $(\mathcal F_n)$;
(ii) for each $n, X_n$ is integrable;
(iii) for each $n, E(X_{n+1} | \mathcal F_n) = X_n$.

Conditions (i) and (ii) are not clear to me ...
We have to check condition (iii), that in our case seems to be :

$P(\varphi|T_i)=P(\varphi)$.


Consider the comment below the formula of page 3 :

By axiom 3, $P(T_0) = 1$ [recall that $T_0 = \emptyset$], so $P(\varphi | T_0) = P(\varphi)$.

Consider now $i=0$; by construction, $T_{i+1} = T_1$ i.e.$= \emptyset \cup \{ \varphi_1$ } or $= \emptyset \cup \{ \lnot \varphi_1 \}$.
Thus, with $P(\varphi|T_i)=P(\varphi \land T_i)/P(T_i)$, due to the fact that we have only two possibility for building $T_1$ : add to $T_0=\emptyset$ either $\varphi_1$ or $\lnot \varphi_1$  we have that [to be verified] :

$P(\varphi|T_1) = P(\varphi|\varphi_1)P(\varphi_1|T_0) + P(\varphi|\lnot \varphi_1)P(\lnot \varphi_1|T_0)$

i.e.

$=P(\varphi_1)/P(T_0) \times P(\varphi \land \varphi_1)/P(\varphi_1) + P(\lnot \varphi_1)/P(T_0) \times P(\varphi \land \lnot \varphi_1)/P(\lnot \varphi_1)$

and simplifying and recalling that $P(T_0)=P(\emptyset)=1$ :

$=P(\varphi \land \varphi_1) + P(\varphi \land \lnot \varphi_1) = P(\varphi)$

by axiom 1.
Iterating this process [see formula page 3], we have that $P(\varphi|T_i)=P(\varphi)$, for all $i$.

Now, applying Ilya's comment above :


$P(\varphi|T_i)$ is a deterministic value, hence it is a martingale iff it is a constant sequence


we have that it is a constant sequence; thus, it is a martingale.


Note
The fact that $P(\emptyset)=1$ means that they treat $\emptyset$ as a tautology.
This fact is consistent with the usual definition of tautology as a formula that is tautologically implied by the empty set of premises :

$\tau$ is a tautology iff $\emptyset \vDash \tau$.

The reason for it is [see Enderton, page 23] :

[Consider $\Sigma \vDash \tau$ and] take the special case in which $\Sigma$ is the empty set $\emptyset$. Observe that it is vacuously true that any truth assignment satisfies every member of $\emptyset$. (How could this fail? Only if there was some unsatisfied member of $\emptyset$, which is absurd.)
Hence we are left with: $\emptyset \vDash \tau$ iff every truth assignment (for the sentence symbols in $\tau$) satisfies $\tau$.

For the same reason, there are no truth assignments "falsifying" $\emptyset$; so, $P(\emptyset)=1$.
