Disjunction and Implication
{{refactor|in progress}} == Theorems == === Modus Tollendo Ponens === {{:Modus Tollendo Ponens/Variant}} === Rule of Material Implication === {{:Rule of Material Implication}} Both of the above come in negative forms: {{begin-eqn}} {{eqn | l = \neg \left({p \implies q}\right)       | o = \dashv \vdash       | r = \neg \left({\neg p \lor q}\right)       | c =  }} {{eqn | l = \neg \left({\neg p \implies q}\right)       | o = \dashv \vdash       | r = \neg \left({p \lor q}\right)       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | ll= \vdash       | l = \left({\neg \left({p \implies q}\right)}\right)       | o = \iff       | r = \left({\neg \left({\neg p \lor q}\right)}\right)       | c =  }} {{eqn | ll= \vdash       | l = \left({\neg \left({\neg p \implies q}\right)}\right)       | o = \iff       | r = \left({\neg \left({p \lor q}\right)}\right)       | c =  }} {{eqn | ll= \vdash       | l = \left({p \lor q}\right)       | o = \iff       | r = \left({\left({p \implies q}\right) \implies q}\right)       | c =  }} {{end-eqn}} They can be seen to be logically equivalent to the forms above. == Proof == {{BeginTableau|\neg \left({p \implies q}\right) \vdash \neg \left({\neg p \lor q}\right)}} {{Premise|1|\neg \left({p \implies q}\right)}} {{Assumption|2|\neg p \lor q|Assume the opposite of what is to be proved ...}} {{SequentIntro|3|2|p \implies q|2|Rule of Material Implication}} {{NonContradiction|4|1, 2|3|1| ... demonstrating a contradiction}} {{Contradiction|5|1|\neg \left({\neg p \lor q}\right)|2|4}}   {{BeginTableau|\neg \left({\neg p \lor q}\right) \vdash \neg \left({p \implies q}\right)}} {{Premise|1|\neg \left({\neg p \lor q}\right)}} {{Assumption|2|p \implies q|Assume the opposite of what is to be proved ...}} {{SequentIntro|3|2|\neg p \lor q|2|Rule of Material Implication}} {{NonContradiction|4|1, 2|3|1| ... demonstrating a contradiction}} {{Contradiction|5|1|\neg \left({p \implies q}\right)|2|4}}   {{LEM|Rule of Material Implication}} {{BeginTableau|\neg \left({\neg p \implies q}\right) \vdash \neg \left({p \lor q}\right)}} {{Premise|1|\neg \left({\neg p \implies q}\right)}} {{Assumption|2|p \lor q|Assume the opposite of what is to be proved ...}} {{SequentIntro|3|2|\neg p \implies q|2|Modus Tollendo Ponens}} {{NonContradiction|4|1, 2|3|1| ... demonstrating a contradiction}} {{Contradiction|5|1|\neg \left({p \lor q}\right)|2|4}}   {{BeginTableau|\neg \left({p \lor q}\right) \vdash \neg \left({\neg p \implies q}\right)}} {{Premise|1|\neg \left({p \lor q}\right)}} {{Assumption|2|\neg p \implies q|Assume the opposite of what is to be proved ...}} {{DeMorgan|3|1|\neg p \land \neg q|1|Conjunction of Negations}} {{Simplification|4|1|\neg p|3|1}} {{ModusPonens|5|1, 2|q|2|4|... from the assumption}} {{Simplification|6|1|\neg q|3|2}} {{NonContradiction|7|1, 2|5|6| ... demonstrating a contradiction}} {{Contradiction|8|1|\neg \left({\neg p \implies q}\right)|2|7}}   {{BeginTableau|p \lor q \vdash (p \implies q) \implies q}} {{Premise|1|p \lor q}} {{Assumption|2|p \implies q}} {{Assumption|3|p}} {{ModusPonens|4|2, 3|q|2|3}} {{Assumption|5|q}} {{ProofByCases|6|2|r|1|3|4|5|5}} {{Implication|7|1|\left({p \implies q}\right) \implies q|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 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 intuitionistic 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 boolean interpretations. $\begin{array}{|cccc||ccccc|} \hline \neg & (p & \lor & q) & \neg & (\neg & p & \implies & q) \\ \hline T & F & F & F & T & T & F & F & F \\ F & F & T & T & F & T & F & T & T \\ F & T & T & F & F & F & T & T & F \\ F & T & T & T & F & F & T & T & T \\ \hline \end{array}$  $\begin{array}{|cccc||ccccc|} \hline \neg & (p & \implies & q) & \neg & (\neg & p & \lor & q) \\ \hline F & F & T & F & F & T & F & T & F \\ F & F & T & T & F & T & F & T & T \\ T & T & F & F & T & F & T & F & F \\ F & T & T & T & F & F & T & T & T \\ \hline \end{array}$