Is this style of proof clear and readable?

Question

$\def\p {\phi} \def\P {\Phi} \def\q {\psi} \def\s {\vDash_{\tiny{PL}}} \def\ns {\nvDash_{\tiny{PL}}} \def\bigc #1{\vec\P_{#1}^\q} \def\pli {\mathscr{I}} \def\val #1{V_\pli(#1)} \def\mf {\p_1,\,\p_2,\,\ldots,\,\p_n} \def\ms #1{\p_{#1}\to (\p_{#1-1}\to (\cdots\to(\p_1\to\q)\cdots))}$

Question

Based on advice received in this question (h/t @MJD), I've rewritten my proof and added in a corollary. My questions are:

• Is it easy to read?

• Does the notation make sense?

• Is there anything that can be improved?

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Proofs

Definitions

1. $\text{wff}:=$ well-formed formula

2. $\p:=$ with, or without, subscript is a wff

3. $\q:=$ with, or without, subscript is a wff

4. $\P_n:=\{\mf\}$ is a non-empty, finite set of $n$ wffs

5. $\bigc {1}:=(\p_1\to\q)$

6. $\bigc {n}:=(\p_n\to\bigc {n-1})$ is a conditional that has been finitely iterated $n$ times (i.e., $\bigc {n}=\ms {n})$

7. $\pli:=$ is a classical propositional logic interpretation

8. $\pli(\cdots):=$ is an interpretation function that assigns truth values to individual sentence letters

9. $\val {\cdots}:=$ is a valuation function that assigns truth values to wffs whose sentence letters have been assigned a truth value by the interpretation function

10. $\val {\p}=\pli (\p)$ if and only if $\p$ is a single sentence letter

11. $\val {\p\to\q}=1$ if and only if $\val {\p}=0$ or $\val {\q}=1$

12. $\val {\P_n}=1$ if and only if each and every $\val {\p\in\P_n}=1$

13. $\val {\bigc {n}}=1$ if and only if there exists a $\val {\p\in\P_n}=0$ or $\val {\q}=1$

14. $\s:=$ is semantic entailment for classic propositional logic

15. $\P_n\s\q$ if and only there exists no $\pli$ such that $\val {\P_n}=1$ and $\val {\q}=0$

16. $\s\bigc {n}$ if and only if there exists no $\pli$ such that $\val {\bigc {n}}=0$

Proof 1

We want to prove that $\P_n\s\q$ if and only if $\s\bigc {n}$. We will do this by showing that both directions of the biconditional hold.

Direction 1

1. For reductio, suppose it's not the case that if $\P_n\s\q$ then $\s\bigc {n}$

2. It follows from line 1 that $\P_n\s\q$ and $\ns\bigc {n}$

3. It follows from line 2 that there exists an $\pli$ such that $\val {\bigc {n}}=0$

4. It follows from line 3 that there exists an $\pli$ such that $\val {\P_n}=1$ and $\val {\q}=0$

5. It follows from line 4 that $\P_n\ns\q$, which contradicts line 2

6. By reductio we've shown if $\P_n\s\q$ then $\s\bigc {n}$

Direction 2

1. For reductio, suppose it's not the case that if $\s\bigc {n}$ then $\P_n\s\q$

2. It follows from line 1 that $\s\bigc {n}$ and $\P_n\ns\q$

3. It follows from line 2 that there exists an $\pli$ such that $\val {\P_n}=1$ and $\val {\q}=0$

4. It follows from line 3 that $\val {\bigc {n}}=0$

5. It follows from line 4 that $\ns\bigc {n}$, which contradicts line 2

6. By reductio we've shown if $\s\bigc {n}$ then $\P_n\s\q$

We've shown that both directions hold, therefore, $\P_n\s\q$ if and only if $\s\bigc {n}$ $\square.$

Corollary

We want to show if $\P_n\s\q$ then $\P_n\cup\{\p_{n+1}\}\s\q$. We will do this via reductio.

Proof 2

1. For reductio, suppose that it's not the case that if $\P_n\s\q$ then $\P_n\cup\{\p_{n+1}\}\s\q$

2. It follows from line 1 that there exists an $\pli$ such that $\P_n\s\q$ and $\P_n\cup\{\p_{n+1}\}\ns\q$, which, from Proof 1, we can express as $\s\bigc {n}$ and $\ns\bigc {n+1}$ respectively

3. It follows from line 2 that $\val {\bigc {n}}=1$

4. It follows from the definitions that $\val{\bigc {n+1}}=\val {\p_{n+1}\to\bigc {n}}$

5. It follows from lines 3 and 4 that $\val {\p_{n+1}\to\bigc {n}}=1$

6. It follows from line 5 that $\s\bigc {n+1}$, hence $\P_n\cup\{\p_{n+1}\}\s\q$, which contradicts line 2

7. By reductio we've shown that if $\P_n\s\q$ then $\P_n\cup\{\p_{n+1}\}\s\q$ $\square$.

Discussion

What these proofs show is that the Deduction Theorem and Weakening for classic propositional logic are valid. The proof of Weakening demonstrates that the addition of an extraneous premise, regardless of $\pli$, can't affect validity. In other words, it's a proof that semantic entailment is monotone. What's quite nice about the proof is that it rests on the familiar valuation function for conditionals.