Happy new year to all!

I have a question related to my Discrete Mathematics examination:

Consider the following (parametrized) propositional formula $$𝜑_𝑛$$:

($$p_1$$V$$p_{2n+1}$$) & (!$$p_1$$V!$$p_2$$) & (!$$p1$$V$$p_2$$V$$p_3$$) & (!$$p_1$$V$$p_2$$V!$$p_3$$V!$$p_4$$) &

& (!$$p_1$$V$$p_2$$V!$$p_3$$V$$p_4$$V$$p_5$$) & ... & (!$$p_1$$V$$p_2$$V!$$p_3$$V$$p_4$$V!$$p_5$$V ... V!$$p_{2n}$$) &

& (!$$p_1$$V$$p_2$$V!$$p_3$$V$$p_4$$V!$$p_5$$V ... V$$p_{2n}$$V$$p_{2n+1}$$)

where 𝑛≥0 and all $$𝑝_1$$, ... $$𝑝_{2𝑛+1}$$ are propositional variables.

Question: For which 𝑛≥0 the formula $$𝜑_𝑛$$ is satisfiable? What method for checking the satisfiability do you use? (-Please name it.) Do you know any other method? (-Please name it.)

I understand how to deal with CNF and DNF, but in this case I do not know that to do with infinity terms, could you provide detailed solution?

Let's see if I understand the pattern correctly.

For $$n=0$$ the formula would be:

$$(p_1 \lor p_1) \land (\neg p_1 \lor p_1)$$

Note that that one is satisfiable by setting $$p_1=1$$

For $$n=1$$ the formula would be:

$$(p_1 \lor p_3) \land (\neg p_1 \lor \neg p_2) \land (\neg p_1 \lor p_2 \lor \neg p_2) \land (\neg p_1 \lor p_2 \lor p_3)$$

That one is also satisfiable: Set $$p_1=1$$ to satisfy the first conjunct, then set $$p_2=0$$ to satisfy the second conjunct (which will also satisfy the third), and finally set $$p_3=1$$ to satisfy the last conjunct.

OK ... starting to maybe see a pattern here, but let's do one more:

For $$n=2$$ the formula would be:

$$(p_1 \lor p_5) \land (\neg p_1 \lor \neg p_2) \land (\neg p_1 \lor p_2 \lor p_3) \land (\neg p_1 \lor p_2 \lor \neg p_3 \lor \neg p_4) \land (\neg p_1 \lor p_2 \lor \neg p_3 \lor p_4 \lor p_5)$$

Again, this is easily satisfied: $$p_1=1$$, $$p_2=0$$, $$p_3=1$$, $$p_4=0$$, and $$p_5=1$$ will satisfy the $$5$$ conjuncts in order.

Well, now the pattern is clear: we can satisfy the expression for any $$n$$, because we can satisfy each of the individual conjuncts in order. That is, we can set the $$i$$-th conjunct true simply by giving $$p_i$$ the correct truth-value. To be specific: we just need to set $$p_{2i-1}=1$$ and $$p_{2i}=0$$ for $$1 \leq i \leq n$$

Question: For which $$n \geq 0$$ the formula $$𝜑_𝑛$$ is satisfiable? What method for checking the satisfiability do you use? (-Please name it.) Do you know any other method? (-Please name it.)

It is satisfiable for all $$n \geq 0$$

The 'method' is simply to find a truth-value assignment for which it is easily shown that it makes all conjuncts, and thus the whole expression, true.