Proof of Theorem by Truth Table

Theorem
Let $\phi$ be a propositional formula whose atoms are $p_1, p_2, \ldots, p_n$.

Let $l$ be the line number of any row in the truth table of $\phi$.

For all $i: 1 \le i \ne n$, let $\hat {p_i}$ be defined as:
 * $\hat {p_i} = \begin{cases}

p_i & : \text {the entry in line } l \text { of } p_i \text { is } \T \\ \neg p_i & : \text {the entry in line } l \text { of } p_i \text { is } \F \end{cases}$

Then:


 * $(1): \quad \hat {p_1}, \hat {p_2}, \ldots, \hat {p_n} \vdash \phi$ is provable if the entry for $\phi$ in line $l$ is $\T$
 * $(2): \quad \hat {p_1}, \hat {p_2}, \ldots, \hat {p_n} \vdash \neg \phi$ is provable if the entry for $\phi$ in line $l$ is $\F$

Proof

 * $(1): \quad$ Suppose $\phi$ is an atom $p$.

Then we need to show that $p \vdash p$ and $\neg p \vdash \neg p$.

These are proved in one line in the proof of the Law of Identity.


 * $(2): \quad$ Suppose $\phi$ is of the form $\neg \phi_1$.

There are two cases to consider:

Suppose $\phi$ evaluates to $\T$.