Propositional Logic - Truth tables - with formulas and given truth values

Question

Qu Morgan Kaufmann. http://myweb.sabanciuniv.edu/rdehkharghani/files/2016/02/The-Morgan-Kaufmann-Series-in-Data-Management-Systems-Jiawei-Han-Micheline-Kamber-Jian-Pei-Data-Mining.-Concepts-and-Techniques-3rd-Edition-Morgan-Kaufmann-2011.pdf Morgan Kaufmann. http://myweb.sabanciuniv.edu/rdehkharghani/files/2016/02/The-Morgan-Kaufmann-Series-in-Data-Management-Systems-Jiawei-Han-Micheline-Kamber-Jian-Pei-Data-Mining.-Concepts-and-Techniques-3rd-Edition-Morgan-Kaufmann-2011.pdf

Answer 1

a.v) $(¬((¬(¬ R)) \vee Q)) \wedge (¬ R)$ I am struggling on this last question, I am getting confused on how to contruct the truth table.

Just substitute the truth values then evaluate the connectives one at a time

$\begin{array}{c:c:c:c:c:c:c:c:c|ll}(\lnot & (\lnot & \lnot & R) & \vee & Q)) & \wedge &\lnot & R&(¬((¬(¬ R)) \vee Q)) \wedge (¬ R) \\ &&&\color{blue}T&&\color{blue}F&&&\color{blue}T &(¬((¬(¬ T)) \vee F)) \wedge (¬ T) \\ &&\color{blue}F&\color{silver}T&&\color{blue}F&&\color{blue}F&\color{silver}T&(¬((¬(F)) \vee F)) \wedge (F) & \tiny\textsf{Hmm...something and false...} \\ &\color{blue}T&\color{silver}F&\color{silver}T&&\color{blue}F&&\color{blue}F&\color{silver}T&(¬((T) \vee F)) \wedge (F) \\ &\color{silver}T&\color{silver}F&\color{silver}T&\color{blue}T&\color{silver}F&&\color{blue}F&\color{silver}T&(¬(T)) \wedge (F) \\ \color{blue}F&\color{silver}T&\color{silver}F&\color{silver}T&\color{silver}T&\color{silver}F&&\color{blue}F&\color{silver}T&(F) \wedge (F) \\ \color{silver}F&\color{silver}T&\color{silver}F&\color{silver}T&\color{silver}T&\color{silver}F&\color{blue}F&\color{silver}F&\color{silver}T&(F) \end{array}$

a.i) $P → Q \land R$ .... I got (False)

$T~\to~T\land F$ is false $\checkmark$

a.ii) P → R / Q .... I got (True)

$T~\to~T\lor F$ is true $\checkmark$

a.iii) (P → Q) → R .... I got (False)

$(T\to F)\to T$ is true. (ex falso (sequitur) quodlibet)

a.iv) ((¬ P) → Q) → (¬ R) .... I got (False)

$((\neg T)\to F)\to(\lnot T)$ is false $\checkmark$