Disjunction and Implication

Theorems
This is sometimes referred to as the disjunctive syllogism or Modus Tollendo Ponens:
 * $$p \or q \dashv \vdash \neg p \implies q$$

This is sometimes referred to as the Rule of Material Implication:
 * $$\neg p \or q \dashv \vdash p \implies q$$

Both of the above come in negative forms:


 * $$\neg \left({p \implies q}\right) \dashv \vdash \neg \left({\neg p \or q}\right)$$
 * $$\neg \left({\neg p \implies q}\right) \dashv \vdash \neg \left({p \or q}\right)$$

Disjunction is definable through implication:


 * $$p\or q \dashv\vdash (p\implies q)\implies q$$

Proof by Natural Deduction
By the tableau method:

Comment
Note that this:


 * $$\neg \left({\neg p \implies q}\right) \dashv \vdash \neg \left({p \or q}\right)$$

can be proved in both directions without resorting to the LEM.

All the others:


 * $$p \or q \vdash \neg p \implies q$$
 * $$\neg p \or q \vdash p \implies q$$
 * $$\neg \left({p \implies q}\right) \vdash \neg \left({\neg p \or q}\right)$$

are not reversible in intuitionist logic.

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

As can be seen by inspection, in all cases the truth values under the main connectives match for all models.

$$\begin{array}{|ccc||cccc|} \hline p & \or & q & \neg & p & \implies & q \\ \hline F & F & F & T & F & F & F \\ F & T & T & T & F & T & T \\ T & T & F & F & T & T & F \\ T & T & T & F & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|cccc||ccccc|} \hline \neg & (p & \or & q) & \neg & (\neg & p & \implies & q) \\ \hline T & F & F & F & T & T & F & F & F \\ F & F & T & T & F & T & F & T & T \\ F & T & T & F & F & F & T & T & F \\ F & T & T & T & F & F & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|ccc||cccc|} \hline p & \implies & q & \neg & p & \or & q \\ \hline F & T & F & T & F & T & F \\ F & T & T & T & F & T & T \\ T & F & F & F & T & F & F \\ T & T & T & F & T & T & T \\ \hline \end{array}$$

$$\begin{array}{|cccc||ccccc|} \hline \neg & (p & \implies & q) & \neg & (\neg & p & \or & q) \\ \hline F & F & T & F & F & T & F & T & F \\ F & F & T & T & F & T & F & T & T \\ T & T & F & F & T & F & T & F & F \\ F & T & T & T & F & F & T & T & T \\ \hline \end{array}$$