# How do I write equivalent formulas using only $\to$ and $\bot$?

for following formulas in propositional logic, how do I write equivalent formulas using only logical symbols → and $$\bot$$:

$$\alpha$$ $$\land$$ $$\beta$$

$$\alpha$$ $$\lor$$ $$\beta$$

$$¬\alpha$$

Is it something like:

$$¬\alpha$$ is equivalent to $$\alpha$$$$\bot$$

$$\alpha$$ $$\lor$$ $$\beta$$ is equivalent to $$¬\alpha$$$$\beta$$ therefore ($$\alpha$$$$\bot$$) → $$\beta$$

$$\alpha$$ $$\land$$ $$\beta$$ is equivalent to $$¬\alpha$$ $$\lor$$ $$¬\beta$$ and is therefore $$\alpha$$$$¬\beta$$ and therefore $$\alpha$$ → ($$\beta$$$$\bot$$)

Also, how do we prove that there is no proposition $$\alpha$$ such that ($$\alpha$$ $$↔$$ ($$¬\beta$$)) is tautology, using only propositional variable $$\alpha$$ and →?

$$\alpha \equiv \alpha \to \bot$$ is correct.
$$\alpha \lor \beta \equiv \neg \alpha \to \beta \equiv (\alpha \to \bot) \to \beta$$ is also correct.
$$\alpha \land \beta \equiv \neg \alpha \lor \neg \beta$$ is incorrect. This owuld mean that at least one out of $$\alpha$$ and $$\beta$$ must be false, but we want none of them to be false. Instead, we want to say "neither not $$\alpha$$ nor not $$\beta$$", so the desired formula is the negation of your statement, $$\neg (\neg \alpha \lor \neg \beta)$$, so the equilalence we end up with is $$(\alpha \to (\beta \to \bot)) \to \bot$$.

There is no propositional variable $$\alpha$$ such that $$(\alpha \leftrightarrow (\neg \beta)).$$

I don't know what is meant by "only using $$\to$$" given that the proof to be presented has to be an informal semantic one" (you can't prove a statement such as "there exists no "\alpha" such that" by using truth tables or logical equivalences), so I assume this was a typo:
With the restriction that $$\alpha$$ be a propositional variable, there are only two cases to distinguish:
1. $$\alpha = \beta$$, i.e. $$\alpha$$ and $$\beta$$ are the same variable, and have the same value under any assignment function. Then obviously, by the definition of $$\neg$$ under no assignment can $$\alpha$$ and $$\neg \beta$$ the same truth value, which contradicts the semantics of $$\leftrightarrow$$ which demands that $$\alpha$$ and $$\beta$$ be true under all the same assignments.
2. $$\alpha \neq \beta$$, i.e. $$\alpha$$ and $$\beta$$ are different propositional variables. Then there exists an assignment under which $$\alpha$$ and $$\beta$$ have the same truth value, which again contradicts the definition of $$\leftrightarrow$$.

• "¬(¬α∨¬β), so the equilalence we end up with is (α→(β→⊥))→⊥." I think you forgot a $\bot$. – Doug Spoonwood Apr 28 at 11:41
• @Doub Spoonwood No, the truth table checks out fine. – lemontree Apr 28 at 21:28
• You're right. I misread the $\lor$ as a $\rightarrow$ I think. – Doug Spoonwood Apr 29 at 18:04