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
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 3, 1
 * ... demonstrating a contradiction ...
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg \left({\neg p \lor q}\right)$
 * Proof by Contradiction
 * 2-4
 * }
 * Proof by Contradiction
 * 2-4
 * }
 * }


 * align="right" | 3 ||
 * align="right" | 2
 * $\neg p \lor q$
 * Sequent Introduction
 * 2
 * Rule of Material Implication
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 3, 1
 * ... demonstrating a contradiction
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg \left({p \implies q}\right)$
 * Proof by Contradiction
 * 2-4
 * }
 * Proof by Contradiction
 * 2-4
 * }
 * }


 * align="right" | 3 ||
 * align="right" | 2
 * $\neg p \implies q$
 * Sequent Introduction
 * 2
 * Modus Tollendo Ponens
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 3, 1
 * ... demonstrating a contradiction
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg \left({p \lor q}\right)$
 * Proof by Contradiction
 * 2-4
 * }
 * Proof by Contradiction
 * 2-4
 * }
 * }


 * align="right" | 3 ||
 * align="right" | 1
 * $\neg p \land \neg q$
 * Sequent Introduction
 * 1
 * De Morgan's Laws: $\neg \left({p \lor q}\right) \vdash \neg p \land \neg q$
 * align="right" | 4 ||
 * align="right" | 1
 * $\neg p$
 * $\land \mathcal E_1$
 * 3
 * align="right" | 5 ||
 * align="right" | 1, 2
 * $\,\!q$
 * $\implies \mathcal E$
 * 2, 4
 * ... from the assumption
 * align="right" | 6 ||
 * align="right" | 1
 * $\neg q$
 * $\land \mathcal E_2$
 * 3
 * align="right" | 7 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 5, 6
 * ... demonstrating a contradiction
 * align="right" | 8 ||
 * align="right" | 1
 * $\neg \left({\neg p \implies q}\right)$
 * Proof by Contradiction
 * 2-7
 * }
 * ... demonstrating a contradiction
 * align="right" | 8 ||
 * align="right" | 1
 * $\neg \left({\neg p \implies q}\right)$
 * Proof by Contradiction
 * 2-7
 * }
 * }
 * }


 * align="right" | 4 ||
 * align="right" | 2,3
 * $q$
 * $\implies \mathcal E$
 * 2,3
 * 2,3


 * align="right" | 6 ||
 * align="right" | 2
 * $q$
 * $\lor \mathcal E$
 * 1, 3 - 4, 5 - 5
 * align="right" | 7 ||
 * align="right" | 1
 * $\left({p \implies q}\right) \implies q$
 * $\implies \mathcal I$
 * 2 - 6
 * }
 * $\implies \mathcal I$
 * 2 - 6
 * }
 * }

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 LEM.

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.

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

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