What could have been the sentences I was thinking about?

Question

I have just thought about an interesting problem in classical propositional logic to propose at this site, but lost my notes... I do recall, however, that the problem involved a particular contingent sentence with three atomic variables, $p$, $q$ and $r$. This first sentence contained exactly two occurrences of connectives, both binary. I am positive that at least one of those connectives was an implication having one of the three aforementioned variables as antecedent.

Question 1. Assuming that the previous information is sufficient to semantically characterize the forgotten sentence, can you determine what was its other connective, as well as the possible logical forms that my sentence could have had?

Now, another thing I do recall is that the problem that I had created consisted in checking that the first sentence above is semantically equivalent to a second sentence, written with the same variables and containing the same connectives as the first sentence, but with one extra implication.

Question 2. What could have been this second sentence? Justify.

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The following extra considerations should be unnecessary to solve the problem:

Word of warning.
While one must not disconsider any information from the above statement when looking for a solution, one should also not read more than what is written. In particular:
(1) Nowhere in the statement of the problem is to be found a claim about the first (or the second) sentence being "syntactically unique". In fact, there is obviously no way, for instance, of avoiding permutation of the variables in a given solution. But there might clearly be many other structurally different sentences (belonging to the same equivalence class!) that solve Question 1. The task is to identify such sentences (preferably without just guessing).
(2) In the statement of the problem, a sentence is supposed to be "semantically characterized" when it is characterized up to logical equivalence --- so, again, you had better think first about what is happening at the level of the quotient algebra, and consider next the syntactic restrictions imposed by the statement of the problem.

Hint.
The assumption made in Question 1 is integral part of the statement, and in fact essential to the existence of a well-determined solution!

Answer 1

Question 1:

p implies NAND(q,r), which is equivalent to (not p) or (not q) or (not r) which is symmetric in the variables. Once you're told that the second gate is a NAND, you know the sentence must have the form _ implies NAND(_,_). Since any permutation of p,r,q in the blanks is equivalent to the symmetric expression given, all possible choices are semantically equivalent. Intuitively, the sentence says that if a variable is true, the other two can't both also be true, which is equivalent to "It can't be the case that all three variables are true"

Answer 2

I have a new answer based on a purely formal logical interpretation of the problem.

Lemma 1: The conditions in paragraph 1 do not specify a syntantically unique sentence.

Proof of lemma: Both the sentences p implies (q or r) and p implies (q and r) satisfying those conditions, as they have three atomic variables p,q,r and two connectives, one an implication implied by a variable. By, they are not logically equivalent as p=True, q=True, r=False satisfies the first but not the second.

Now, afterwards, we're asked to assume that the previous information is sufficient to semantically characterize the forgotten sentence, which is the negation of Lemma 1. For having both Lemma 1 and its negation, we use the principle of explosion to derive all statements.

Therefore, we derive that the forgotten statement was "I am the eggman, I am the walrus." As to the question of "the possible logical forms that my sentence could have had", it is the set of logical forms.

Answer 3

I'll have a go with :

$p \rightarrow (q \leftrightarrow r)$

together with

$(p \rightarrow q) \leftrightarrow (p \rightarrow r)$

This reads as "p implies (q if and only if r)", and "(p implies q) if and only if (p implies r).

Answer 4

EDIT: As J Marcos points out in the comments, this answer cannot possibly be correct, as the 'lost sentence' it gives is tautological, not contingent as specified in the question.

It's been a while since I've done anything with formal logic, so please forgive (and correct) me if I butcher the terminology. I have only attempted to answer question 1.

Based on the information given there are three possible forms (up to relabeling of variables) for the forgotten sentence.

1. $p \rightarrow (q \text{ B } r)$
2. $(p \rightarrow q) \text{ B } r$
3. $p \text{ B }(q\rightarrow r)$

where B is some binary connective. Note that I left out $(p \text{ B } q) \rightarrow r$, as the implication doesn't have an atomic antecedent (unless B is also implication, in which case we're back in case 2).

There is no other information in the question that would allow us to choose any particular one of these forms over the others. Thus, the problem is only well-defined* if there is exactly one choice of B such that the three forms are semantically equivalent. As it happens, there is, and we can derive it as follows.

When $p$ is false, the three forms have the values $True$, $True\text{ B }r$, and $False\text{ B }(q\rightarrow r)$, respectively. These must all be equal regardless of the value of $r$, so $True\text{ B } True = True\text{ B }False = True$.

When $p$ is true and $q$ is false, form 1 reduces to $True \text{ B } True = True$ and form 2 to $False \text{ B } r$. Again these must be equal for either value of $r$, so $False \text{ B } True = False \text{ B } False = True$.

This is enough to completely characterize B: it is the binary connective that always yields $True$ regardless of its operands.

*I'm cheating a bit here, I think -- J Marcos only assumes that the information is enough to define the sentence up to semantic equivalence, which is not quite the same as assuming the question "what was the other connective" has a well-defined answer. To make up the gap I'd need to show that semantic equivalence of $p \rightarrow (q \text{ C } r)$, $(p\rightarrow q)\text{ D } r$, and $p\text{ E }(q\rightarrow r)$ is enough to imply $\text{ C } = \text{ D } = \text{ E } = \text{ B }$, where B is the connective derived above. Proving this is completely mechanical (write down the truth tables for C, D, and E; plug in different values of $p$, $q$, and $r$ to 1, 2, and 3; and follow your nose) but would be tedious to write up, so I leave it as an exercise to the reader.