Disjunction and Implication

Rule of Material Implication
Both of the above come in negative forms:

Disjunction is definable through implication:


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

Alternative rendition
They can alternatively be rendered as:

They can be seen to be logically equivalent to the forms above.

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p \implies q$
 * Sequent Introduction
 * 2
 * Rule of Material Implication
 * Rule of Material Implication


 * align="right" | 3 ||
 * align="right" | 2
 * $\neg p \lor q$
 * Sequent Introduction
 * 2
 * Rule of Material Implication
 * Rule of Material Implication


 * align="right" | 3 ||
 * align="right" | 2
 * $\neg p \implies q$
 * Sequent Introduction
 * 2
 * Modus Tollendo Ponens
 * Modus Tollendo Ponens


 * align="right" | 3 ||
 * align="right" | 1
 * $\neg p \land \neg q$
 * Sequent Introduction
 * 1
 * De Morgan's Laws: $\neg \left({p \lor q}\right) \vdash \neg p \land \neg q$
 * De Morgan's Laws: $\neg \left({p \lor q}\right) \vdash \neg p \land \neg q$

Comment
Note that this:


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

can be proved in both directions without resorting to the Law of Excluded Middle.

All the others:


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

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.

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

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