Conjunction of Disjunctions with Complements implies Disjunction

Theorem

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

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $p \lor r$
 * $\land$ elimination $1$
 * 1
 * The aim is to use $\lor$ elimination on this ...
 * The aim is to use $\lor$ elimination on this ...


 * align="right" | 4 ||
 * align="right" | 3
 * $p \lor q$
 * $\lor$ introduction $1$
 * 3
 * ... and demonstrate the conclusion
 * align="right" | 5 ||
 * align="right" | 1
 * $q \lor \neg r$
 * $\land$ elimination $2$
 * 1
 * align="right" | 6 ||
 * align="right" | 1
 * $\neg r \lor q$
 * Rule of Commutation
 * 5
 * align="right" | 7 ||
 * align="right" | 1
 * $r \implies q$
 * Rule of Material Implication
 * 6
 * align="right" | 7 ||
 * align="right" | 1
 * $r \implies q$
 * Rule of Material Implication
 * 6
 * 6


 * align="right" | 9 ||
 * align="right" | 8, 1
 * $q$
 * Modus Ponendo Ponens
 * 7, 8
 * align="right" | 10 ||
 * align="right" | 8, 1
 * $p \lor q$
 * $\lor$ introduction $2$
 * 9
 * ... and demonstrate the conclusion.
 * align="right" | 11 ||
 * align="right" | 1
 * $p \lor q$
 * $\lor$ elimination
 * 2, 3-4, 8-10
 * }
 * $p \lor q$
 * $\lor$ elimination
 * 2, 3-4, 8-10
 * }
 * }