# Prove the soundness of propositional logic without using induction?

I want to prove the soundness of propositional logic without using induction. I think I can do that via a process that's basically universal introduction (i.e., demonstrate something about an arbitrary member and infer that it applies universally). As an example of this approach I've picked a new inference rule to prove.

#### My questions:

• Is my approach valid?
• Is my proof correct?
• Is it easy to understand and follow?
• Do I need to add more/less detail?
• How else can I improve it?

#### Proof

We want to show that $$\boxed{\dfrac{\Gamma_1 ,\,\phi\vdash\psi\quad\Gamma_2,\,\lnot\phi\vdash\psi}{\Gamma_1 ,\,\Gamma_2\vdash\psi}}\def\pa {((\Gamma_1\land\phi)\longrightarrow\psi)} \def\pb {((\Gamma_2\land\lnot\phi)\longrightarrow\psi)} \def\ca {((\Gamma_1\land\Gamma_2)\longrightarrow\psi)} \def\lto {\longrightarrow} \def\val#1{V_\mathscr{I}( #1 )} \def\pli {\text{PL-interpretation, \mathscr{I}},} \def\inp{\mathscr{I}}$$ is a truth preserving inference rule without using induction. To do so we'll convert the rule to an axiom schema and show that it's valid.

#### Conversion Rules

##### Symbols
• "," and the "$$\quad$$" convert to conjunction
• "$$\vdash$$" and the vinculum convert to implication
• $$\Gamma$$, with or without subscript, is a finite set of conjoined wffs, so it's simply a wff with it's own valuation rules
##### Valuation of $$\Gamma$$
• $$\val{\Gamma}=1$$ iff, for all $$\gamma\in\Gamma$$, it's the case that $$\val{\gamma}=1$$ and $$\Gamma\neq\emptyset$$

#### Conversion

• $$\Gamma_1 ,\,\phi\vdash\psi:= \pa$$

• $$\Gamma_2,\,\lnot\phi\vdash\psi:= \pb$$

• $$\Gamma_1 ,\,\Gamma_2\vdash\psi:= \ca$$

• Putting it all together, the rule converts to - $$\boxed{((\pa\land\pb)\lto\ca)}$$

#### Proof that the axiom schema is valid

1. Assume for reductio that $$\val {((\pa\land\pb)\lto\ca)}=0$$

2. It follows from (1) that $$\val {\ca}=0$$

3. It follows from (2) that $$\val{\Gamma_1}=1$$, $$\val{\Gamma_2}=1$$, and $$\val{\psi}=0$$

4. It follows from (1) that $$\val{\pa}=1$$

5. It follows from (3) and (4) that $$\val{\phi}=0$$

6. It follows from (5) that $$\val{\lnot\phi}=1$$

7. It follows from (1) that $$\val{\pb}=1$$

8. It follows from (3) and (6) that $$\val{\pb}=0$$, which contradicts (7) $$\boxed{}$$

##### Example Proof

$$\begin{array}{lrcll} 1.&\phi&\vdash &\phi &\text{RA}\\ 2.&\phi&\vdash &\lnot\psi\lor\phi &\text{1, \lorI}\\ 3.&\phi&\vdash &\psi\to\phi &\text{2, Abbrv}\\ 4.&\phi&\vdash & (\phi\to\psi)\lor(\psi\to\phi) &\text{3, \lor I}\\ 5.&\lnot\phi&\vdash &\lnot\phi &\text{RA}\\ 6.&\lnot\phi&\vdash &\lnot\phi\lor\psi &\text{5, \lorI}\\ 7.&\lnot\phi&\vdash &\phi\to\psi &\text{6, Abbrv}\\ 8.&\lnot\phi&\vdash & (\phi\to\psi)\lor(\psi\to\phi) &\text{7, \lor I}\\ 9.&\emptyset &\vdash & (\phi\to\psi)\lor(\psi\to\phi) &\text{4, 8 New Rule}\\ \end{array}$$

• ... provided that $\Gamma_i$ are finite sets of formulas, otherwise $\Gamma_i \land \phi$ is not a formula. Commented Sep 10, 2021 at 10:00
• @Mauro ALLEGRANZA, I've explicitly mentioned that $\Gamma$ is a finite set, but I could draw further attention to it if it makes things clearer? Commented Sep 10, 2021 at 10:07
• But this is the key-point: if the number of cases is finite, we do not need induction. It is enough to check them one-by-one. Commented Sep 10, 2021 at 10:09
• @Mauro ALLEGRANZA, while each substitution instance is composed of a finite number of wffs, there are an infinite number of substitution instances, though Commented Sep 10, 2021 at 10:18
• To prove your concerned sentential soundness theorem, normally we need to use RAA to prove each logical connective's soundness case by case. Since each connective has different meaning, what's the most important principle upon which you can claim your single universal inference rule can ensure each row of the truth table of every connective to acting exactly according to their respective definition? The famous XOR connective, for example, has a very peculiar elimination rule, how your proposed theory can ensure it's not acting like the usual wrong elimination rule mimic disjunction syllogism? Commented Sep 10, 2021 at 23:35