Q:

Bayesian inference: updating under uncertainty

Bayes Theorem:
$$Pr(A|B) = \frac{Pr(A)Pr(B|A)}{Pr(B)} \tag{0}$$
Furthermore:
$$Pr(A) = Pr(A|B)Pr(B) + Pr(A|\neg B)Pr(|\neg B) \tag{1}$$
Scenario:
Suppose we are not given that an event $B$ has occurred. Assuming we are given a base Probability of $A$:  $(Pr(A_1)$ (this is our prior probability), subsequently, we are now given a $Pr(A|B_1)$, a $Pr(B_1)$, a $Pr(A|\neg B_1)$, and a $Pr(\neg B_1)$ as additional information. Assuming, that I now want to calculate a $Pr(A_2)$ that is the posterior probability in light of the new information.   
From $(1)$:
$$Pr(A_2) = Pr(A|B_1) Pr(B_1) + Pr(A|\neg B_1) Pr(\neg B_1)\tag{2}$$  
From $(0)$:
$$Pr(A|B_1) = \frac{Pr(A_1)Pr(B_1|A)}{Pr(B_1)} \qquad :\qquad Pr(A|\neg B_1) = \frac{Pr(A_1)Pr(\neg B_1|A)}{Pr(\neg B_1)} \tag{3}$$   
Rewriting $(2)$ in light of $(3)$:
$$Pr(A_2) = \frac{Pr(A_1)Pr(B_1|A)}{Pr(B_1)} \times Pr(B_1) + \frac{Pr(A_1)Pr(\neg B_1|A)}{Pr(\neg B_1)} \times Pr(\neg B_1)$$   
Which gives us:
$$Pr(A_2) = Pr(A_1)Pr(B_1|A) + Pr(A_1)Pr(\neg B_1|A)$$
Simplifying to:
$$Pr(A_2) = Pr(A_1)\left(Pr(B_1|A) +Pr(\neg B_1|A)\right)\tag{4}$$    
Generalising from $(4)$:
$$Pr(A_{l+1}) = Pr(A_l)\left(Pr(B_l|A) +Pr(\neg B_l|A)\right)\tag{5}$$    
My Question: 

Is $(5)$ valid Bayesian inference? 
If not, how do I perform valid inference in the specified scenario?

A:

Your equations do not allow one to make any kind of inference (Bayesian or not):  they just state that (in your notation) $Pr(A_{l+1})=Pr(A_l)=Pr(A_1)$. Proof:
$$Pr(A_{l+1}) = Pr(A_l)\left(Pr(B_l|A) +Pr(\neg B_l|A)\right)\tag{5}$$  
Now, conditional probabilities are probabilities, which means that they follow all the usual probability rules. In particular, since $B_l$ and $\neg B_l$ are mutually exclusive events, we have
$$Pr(B_l|A) +Pr(\neg B_l|A)=Pr(B_l \cup \neg B_l|A)=Pr(\Omega|A)=1$$
Thus 
$$Pr(A_{l+1}) = Pr(A_l) \cdot 1 = Pr(A_l) = Pr(A_1) $$ Q.E.D. 

