Question about conjunctions

Question

Would this principle be true for conjunctions:

For any proposition p, if p is a conjunction with at least three conjuncts, then there are more distinct conjunctions that can be created out of the conjuncts of p than there are conjuncts of p.

Presumably for any conjunction with only two conjuncts, i.e., (P & Q), the only conjunction that can be created out of its conjuncts is (P & Q). So it's not the case that there are more distinct conjunctions that can be created out of the conjuncts of (P & Q) than there are conjuncts of (P & Q).

But it seems that for any conjunction with at least three conjuncts, e.g., (P & Q & R) or (P & Q & R & S), the principle in question would hold true. At least for finite conjunctions with three or more conjuncts, i.e., conjunctions with a finite number of conjuncts where the number of conjuncts is three or more, we know that this principle would hold true, since for such conjunctions we can just use conjunction elimination on them to single out each of their conjuncts and then we can use conjunction introduction on these conjuncts to form all possible conjunctions (and the number of possible conjunctions we'd be able to form would be larger than the number of conjuncts we used in question).

Would this principle, however, hold for infinite conjunctions?

Answer 1

Suppose we consider the conjunction P & Q.

We could create many conjunctions from the conjuncts P and Q. For example, there is P & ¬⊥. The "¬⊥" means "not false" or "true". And if that is acceptable, then there is also this conjunction P & ¬⊥ & ¬⊥. This could go on indefinitely and we haven't even considered Q. We could theoretically make a countably infinite number of conjunctions using this method on the conjuncts of any proposition composed of conjuncts.

Consider the principle:

For any proposition p, if p is a conjunction with at least three conjuncts, then there are more distinct conjunctions that can be created out of the conjuncts of p than there are conjuncts of p.

Given the above construction to use "& ¬⊥" over and over again, even if there are only two conjuncts, one could construct more than two distinct conjunctions. Even with an atomic proposition P one could form an indefinite number of conjunctions by doing the same to it.

If we have a countably infinite number of conjuncts each making an infinite number of conjunctions one could create a infinite number of conjunctions. A diagonal argument like the one Cantor used would likely show this was an uncountably infinite number of conjuncts.