Chapter 2 - The Logic of Compound Statements - Exercise Set 2.1 - Page 38: 41

contradiction

$(p\land \lnot q)\land(\lnot p \lor q)$ $\underline{\ p\ |\ q\ |\ \lnot p\ |\ \lnot q\ |\ \ \ (p\land \lnot q)\land(\lnot p \lor q)\ \ \ |result}$ $T\ |\ T|\ \ \ F\ |\ \ F\ |\ \ \ \ (T\land F)\land(F \lor T)\ \ \ |\ \ \ \ F$ $T\ |\ F|\ \ \ F\ |\ \ T\ |\ \ \ \ (T\land T)\land(F \lor F)\ \ \ |\ \ \ \ F$ $F\ |\ T|\ \ \ T\ |\ \ F\ |\ \ \ \ (F\land F)\land(T \lor T)\ \ \ |\ \ \ \ F$ $F\ |\ F|\ \ \ T\ |\ \ T\ |\ \ \ \ (F\land T)\land(T \lor F)\ \ \ |\ \ \ \ F$