Tautology and Contradiction

Theorems
A contradiction implies and is implied by the negation of a tautology:


 * $$\bot \dashv \vdash \neg \top$$

That is, a falsehood can not be true, and a non-truth is a falsehood.

A tautology implies and is implied by the negation of a contradiction:


 * $$\top \dashv \vdash \neg \bot$$

That is, a truth can not be false, and a non-falsehood must be a truth.

A conjunction with a tautology:


 * $$p \and \top \dashv \vdash p$$

A disjunction with a tautology:


 * $$p \or \top \dashv \vdash \top$$

A conjunction with a contradiction:


 * $$p \and \bot \dashv \vdash \bot$$

A disjunction with a contradiction:


 * $$p \or \bot \dashv \vdash p$$

Proof by Natural deduction
These are proved by the Tableau method.


 * $$\bot \vdash \neg \top$$:


 * $$\neg \top \vdash \bot$$:


 * $$\top \vdash \neg \bot $$:


 * $$\neg \bot \vdash \top$$:


 * $$p \and \top \vdash p$$:


 * $$p \vdash p \and \top$$:


 * $$p \or \top \vdash \top$$:


 * $$\top \vdash p \or \top$$:


 * $$p \and \bot \vdash \bot$$:


 * $$\bot \vdash p \and \bot$$:


 * $$p \or \bot \vdash p$$:


 * $$p \vdash p \or \bot$$:

Comment
Note that the proofs of:


 * $$\neg \bot \vdash \top$$
 * $$\neg \top \vdash \bot$$
 * $$p \vdash p \and \top$$
 * $$p \or \top \vdash \top$$

rely (directly or indirectly) upon the Law of the Excluded Middle - and it can be seen that they are just another way of stating that truth.

"If it's not false, it must be true" and "If it's not true, it must be false" are indeed valid only in the context where there are only two truth values. From the intuitionist perspective, these results do not hold.

Proof by Truth Table
We apply the Method of Truth Tables to the propositions.

As can be seen by inspection, in each case, the truth values in the appropriate columns match for all models.

$$\begin{array}{|c|cc||c|cc|} \hline \top & \neg & \bot & \bot & \neg & \top \\ \hline T & T & F & F & F & T \\ \hline \end{array}$$

$$\begin{array}{|c|ccc||c|ccc|} \hline p & p & \and & \top & \top & p & \or & \top \\ \hline F & F & F & T & T & F & T & T \\ T & T & T & T & T & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|c|ccc||c|ccc|} \hline \bot & p & \and & \bot & p & p & \or & \bot \\ \hline F & F & F & F & F & F & F & F \\ F & T & F & F & T & T & T & F \\ \hline \end{array}$$

Proof by Boolean Interpretation
Let $$p$$ be a logical formula.

Let $$v$$ be any arbitrary boolean interpretation of $$p$$.

Then $$v \left({p}\right) = T \iff v \left({\neg p}\right) = F$$ by the definition of the logical not.

Since $$v$$ is arbitrary, $$p$$ is true in all interpretations iff $$\neg p$$ is false in all interpretations.

Similarly, $$v \left({p}\right) = F \iff v \left({\neg p}\right) = T$$ by the definition of the logical not.

Since $$v$$ is arbitrary, $$p$$ is false in all interpretations iff $$\neg p$$ is true in all interpretations.

Hence $$\bot \dashv \vdash \neg \top$$ and $$\top \dashv \vdash \neg \bot$$.