#### Question

A lot of things I'm trying to prove just now are turning into "notational hell", which I think makes them very hard to read. I've tried to cut down on this by assuming my reader will understand what definitions are in play, modularising my proofs and skipping explanation of steps that I hope are obvious. I've also tried relabeling formulae with short names (i.e., $\def\val#1{V_\pli(#1)}\def\p{\phi}\def\q{\psi}\def\s{\vDash_{\tiny\text{PL}}}\def\ns{\nvDash_{\tiny\text{PL}}}\def\pli{\mathscr{I}}\def\aa{\p_1,\,\p_2,\,\ldots,\,\p_n\s\q}\def\ab{\p_1,\,\p_2,\,\ldots,\,\p_n\ns\q}\def\ba{\s\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))}\def\bb{\ns\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))}\s\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots)):=\,\s Q),$ but for proofs of any length it seems to be more confusing than helpful. **How do I make proofs more readable without sacrificing clarity?**

#### Example Proof

Let $\p$ and $\q$ be wffs and $n\in\mathbb{N}$ (please note that $0\not\in\mathbb{N}$). We want to show $\def\p{\phi}\def\q{\psi}\def\s{\vDash_{\tiny\text{PL}}}\def\ns{\nvDash_{\tiny\text{PL}}}\p_1,\,\p_2,\,\ldots,\,\p_n\s\q$ iff $\s\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))$.

In (L1) I prove both directions of the biconditional, which I don't think I need to do because we're dealing with "=" - is this correct? I also think that (L1) is so basic that "by inspection" is appropriate - is that fair?

##### Lemma 1 (L1)

We want to show by induction that for some PL-interpretation, $\pli,$ $\val{\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))}=0$ iff $\val{\p_n}=\val{\p_{n-1}}=\cdots=\val{\p_1}=1$ and $\val{\q}=0$.

*Base Case*

- If $\val{\p\to\q}=0$ then, by definition, $\val{\p}=1$ and $\val{\q}=0$. If $\val{\p}=1$ and $\val{\q}=0$, then, by definition, $\val{\p\to\q}=0$

*Induction Hypothesis (IH)*

- Assume for some arbitrary $k\in\mathbb{N}$ that $\val{p_k\to(\p_{k-1}\to(\cdots\to(\p_1\to\q)\cdots))}=0$ and $\val{\p_k}=\val{\p_{k-1}}=\cdots=\val{\p_1}=1$ and $\val{\q}=0$

*Induction Step*

If $\val{\p_{k+1}\to(p_k\to(\p_{k-1}\to(\cdots\to(\p_1\to\q)\cdots)))}=0,$ then, as we know $\val{p_k\to(\p_{k-1}\to(\cdots\to(\p_1\to\q)\cdots))}=0$ from the (IH), $\val{\p_{k+1}}=1$. From the (IH) $\val{\p_k}=\val{\p_{k-1}}=\cdots=\val{\p_1}=1$ and $\val{\q}=0$, thus $\val{\p_{k+1}}=\val{\p_k}=\val{\p_{k-1}}=\cdots=\val{\p_1}=1$ and $\val{\q}=0$

Let $\val{\p_{k+1}}=1$. From the (IH) $\val{\p_k}=\val{\p_{k-1}}=\cdots=\val{\p_1}=1,$ $\val{\q}=0,$ and $\val{p_k\to(\p_{k-1}\to(\cdots\to(\p_1\to\q)\cdots))}=0,$ thus $\val{\p_{k+1}\to(p_k\to(\p_{k-1}\to(\cdots\to(\p_1\to\q)\cdots)))}=0$

##### Proof of First Direction (P1)

For reductio, suppose it is not the case that $\aa\implies\ba$

It follows from (1) that there exists an $\def\pli{\mathscr{I}}\pli$ such that $\def\aa{\p_1,\,\p_2,\,\ldots,\,\p_n\s\q}\def\ab{\p_1,\,\p_2,\,\ldots,\,\p_n\ns\q}\def\ba{\s\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))}\def\bb{\ns\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))}\aa$ and $\bb$

It follows from (2) that $\def\val#1{V_\pli(#1)}\val{\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))}=0$

From (L1), the valuation on (3) can only occur when $\val{\p_n}=\val{\p_{n-1}}=\cdots=\val{\p_1}=1$ and $\val{\q}=0$

It follows from (4) that $\ab$, which contradicts (2) and proves our first direction

##### Proof of Second Direction (P2)

For reductio, suppose it's not the case that $\ba\implies\aa$

It follows from (1) that there exists an $\pli$ such that $\ba\text{ and }\ab$

From (2) we have $\ab$, thus, $\val{\p_n}=\val{\p_{n-1}}=\cdots=\val{\p_1}=1$ and $\val{\q}=0$

It follows from (3) and (L1) that $\bb\text{,}$ which contradicts (2) and proves our second direction

(P1) and (P2) prove both directions of the biconditional, hence $\def\p{\phi}\def\q{\psi}\def\s{\vDash_{\tiny\text{PL}}}\def\ns{\nvDash_{\tiny\text{PL}}}\p_1,\,\p_2,\,\ldots,\,\p_n\s\q$ iff $\s\p_n\to(\p_{n-1}\to(\cdots\to(\p_1\to\q)\cdots))\,\square.$