You shouldn't be assigning propositional variables to the pieces of the sentences; they should be predicates, so that you can substitute the day of the week.
- "If I take the day off, it either rains or snows."
Let's call this one $$(\forall d)(T (d) \to (R(d) \vee s(d)),$$
where $T(d)$ is "I take take (day) $d$ off", $R(d)$ is "It rains on (day) $d$", and $S(d)$ is "It snows on (day) $d$".
- "I took Tuesday off or I took Thursday off."
This now becomes $T({\rm Tuesday}) \vee T({\rm Thursday})$.
- "It was sunny on Tuesday."
This becomes $S({\rm Tuesday})$.
- "It did not snow on Thursday."
This becomes $\neg s({\rm Thursday})$. Now you can use the rules of inference (including the ones for quantifiers).
(1) $(\forall d)(T (d) \to (R(d) \vee s(d))$ [Assumption]
(2) $T({\rm Tuesday}) \vee T({\rm Thursday})$ [Assumption]
(3) $S({\rm Tuesday})$ [Assumption]
(4) $\neg s({\rm Thursday})$ [Assumption]
We're also missing some key information. It cannot be both rainy and sunny on day $d$, so we have
(5) $(\forall d)(\neg R(d) \vee \neg S(d))$ [Assumption]
Similarly, it cannot be sunny and snow at the same time. Thus
(6) $(\forall d)(\neg s(d) \vee \neg S(d))$ [Assumption]
Now we can use some rules of inference. It was sunny on Tuesday, so we show that it wasn't raining or snowing on Tuesday.
(7) $\neg R({\rm Tuesday}) \vee \neg S({\rm Tuesday})~~~~$ [Universal Instantiation, (5)]
(8) $\neg R({\rm Tuesday})~~~~$ [Disjunctive Syllogism, (7), (3)]
(9) $\neg s({\rm Tuesday}) \vee \neg S({\rm Tuesday})~~~~$ [Universal Instantiation, (6)]
(10) $\neg s({\rm Tuesday})~~~~$ [Disjunctive Syllogism, (9), (3)]
(11) $\neg R({\rm Tuesday}) \wedge \neg s({\rm Tuesday})~~~~$ [Conjunction, (8), (10)]
(12) $\neg (R({\rm Tuesday}) \vee s({\rm Tuesday}))~~~~$ [Logically equivalent to (11)]
Now we show what days you took off.
(13) $T({\rm Tuesday}) \to (R({\rm Tuesday}) \vee s({\rm Tuesday}))~~~~$ [Universal Instantiation, (1)]
(14) $\neg T({\rm Tuesday})~~~~$ [Modus Tollens, (12), (13)]
(15) $T({\rm Thursday})~~~~$ [Disjunctive Syllogism, (14), (2)]
Now you know you took Thursday off. Now to determine the weather on Thursday.
(16) $T({\rm Thursday}) \to (R({\rm Thursday}) \vee s({\rm Thursday}))~~~~$ [Universal Instantiation, (1)]
(17) $R({\rm Thursday}) \vee s({\rm Thursday})~~~~$ [Modus Ponens, (16), (15)]
(18) $R({\rm Thursday})~~~~$ [Disjunctive Inference, (17), (4)]
Thus it rained on Thursday.
As @MauroALLEGRANZA rightly points out, reading a textbook dedicated to the subject (rather than one that necessarily glosses over it) is the answer my silly question deserves.
Be that as it may, it is perhaps not the worst idea to clarify my understanding of the matter.
Edit: The notation under discussion (alongside summation and the like) is an example of an iterated binary operation, described here.
Based on the (very helpful) comments by @MauroALLEGRANZA and @DaveL.Renfro, this notation harks to sigma and pi notation for summation and product operations, respectively. So, working from the inside out:
$$\bigwedge_{i=1}^{2} \bigwedge_{n=1}^{2} \bigvee_{j=1}^{2} p(i,j,n) \equiv \bigwedge_{i=1}^{2} \bigg(\bigwedge_{n=1}^{2} \bigg(\bigvee_{j=1}^{2} p(i,j,n) \bigg) \bigg)$$
With that in mind, the following:
\begin{align}
\bigwedge_{i=1}^2 \bigwedge_{n=1}^2 \bigvee_{j=1}^2 p(i,j,n) & \equiv \bigwedge_{i=1}^2 \bigwedge_{n=1}^2 p(i,1,n) \vee p(i,2,n) \\
& \equiv \bigwedge_{i=1}^2 [p(i,1,1) \vee p(i,2,1)] \wedge [p(i,1,2) \vee p(i,2,2)] \\
& \equiv [p(1,1,1) \vee p(1,2,1)] \wedge [p(1,1,2) \vee p(1,2,2)] \\
& \wedge [p(2,1,1) \vee p(2,2,1)] \wedge [p(2,1,2) \vee p(2,2,2)]
\end{align}
This final equivalence I understand to be in conjunctive normal form, which is to say a sequence of disjunctive clauses connected by conjunction.
Clearly I have a long, long way to go with all of this. Many thanks in any event to those who took the time to help me understand.
Best Answer
They look like "big and" signs.
That is, they are the general conjunction of all the statements under them. Here is a reference link.