# Can this disjunction be eliminated?

This is a question about classical propositional logic.

Definitions: If there is a proof from $$\alpha$$ to $$\beta$$, we'll write $$\alpha \vdash \beta$$. We'll say that $$\alpha$$ is equivalent to $$\beta$$ if $$\alpha \vdash \beta$$ and $$\beta \vdash \alpha$$ and I'll write it $$\alpha \equiv \beta$$. A literal is a propositional variable or its negation. I'll say that a formula is in conjunctive form if it can be written as a conjunction of literals, for example: $$a \land b \land \lnot c$$. A disagreement is a case where two formulas $$\alpha$$ and $$\beta$$ are in conjunctive form and have a propositional variable such that one is positive in one formula and negative in the other. So for example, the formulas $$b \land c$$ and $$\lnot b \land c$$ disagree on the variable $$b$$.

First, we observe that in cases where $$\alpha$$ and $$\beta$$ have a single disagreement, their disjunction may be equivalent to another formula in conjunctive form. For example,

$$((a \land b \land c) \lor (a \land b \land \lnot c)) \equiv a \land b$$

Question: Consider three conjunctive forms $$\alpha$$, $$\beta$$, and $$\varphi$$. The formulas $$\alpha$$ and $$\beta$$ are over the same propositional variables and have 2 or more disagreements. Is the following ever possible?

$$(\alpha \lor \beta) \equiv \varphi$$

I believe no such $$\varphi$$ exists, but my attempts to prove this seem laborious. Is there a short proof of this?

Write $$\alpha = a_1 \land a_2 \land \cdots \land a_n$$, $$\beta = b_1 \land b_2 \land \cdots \land b_n$$. By properties of $$\land$$ we may assume $$a_i = b_i$$ or $$a_i = \neg b_i$$ for all $$i$$ and $$a_1 = \neg b_1, a_2 = \neg b_2$$. We now treat $$a_1, a_2, \dots, a_n$$ as free variables. Set $$\varphi := \alpha \lor \beta$$ and assume it is a conjuction. Note that $$\varphi$$ then consists of the same free variables $$a_1, a_2, \dots, a_n$$ (so it is a conjuction of some subset of these). Now, we can either construct a truth table for $$\varphi$$, or see directly from the formula that $$\varphi = 1$$ iff $$a_1 = a_2 = \cdots = a_n = 1$$ or $$b_1 = b_2 = \cdots = b_n = 1$$, so the truth table has exactly two rows for which $$\varphi = 1$$.
As $$\varphi$$ is a conjuction, this is only possible if it is independent of exacty one of the variables $$a_1, a_2, \dots, a_n$$. By this I mean that $$\varphi$$ is a conjuction of some $$n - 1$$ of these variables or their negations, e.g. perhaps $$\varphi = a_1 \land \neg a_2 \land a_3 \land \neg a_4 \land \cdots \land a_{n - 1}$$, but there are definitely exactly $$n - 1$$ of them. Really, if there were more, then there'd be exactly $$n$$, so $$\varphi$$ would be true in exactly one row. If there were fewer, we could assume that $$\varphi$$ is independent of $$a_k, a_{k + 1}, \dots, a_{n - 1}, a_n$$. Then it is a conjuction of the first $$k - 1$$ terms and, as a conjuction, it is true in exactly one interpretation of those terms. But the $$n - k + 1$$ leftover terms then allow for $$2^{n - k + 1}$$ different "repetitions" of the interpretation which makes $$\varphi$$ true in the truth table, making it true for many more than $$2$$ rows.
So, $$\varphi$$ depends on exactly $$n - 1$$ variables. One of these is either $$a_1$$ or $$a_2$$, without loss of generality $$a_1$$. But then $$\varphi$$ is true both when $$a_1 = 1$$ and when $$a_1 = 0$$, which isn't possible for a conjunction. Contradiction.