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can you help me solving this problem, I got stucked answering it in the beginning and hoping that you can help me with this. Thanks!

~Q ∧ (P→Q) ∧ (R ∨ ~Q)

Shaun
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JDA
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    $\lnot Q \land (P \rightarrow Q)$ is equivalent to $\lnot P$. Also $\lnot Q \land (R \lor \lnot Q)$ is equivalent to $\lnot Q$. You are hence left with $\lnot P$. – Bermudes Apr 02 '19 at 07:03
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    What means "simplification"? DNF? (https://en.wikipedia.org/wiki/Disjunctive_normal_form) Another thing? – Martín-Blas Pérez Pinilla Apr 02 '19 at 07:04
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    @Bermudes $\neg Q\wedge(P\rightarrow Q)$ is not equivalent to $\neg P$, it just implies $\neg P$. Hence, we are not "left" with only $\neg P$. – Casper Apr 03 '19 at 07:12

2 Answers2

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If we go by truth table as follows:

\begin{array} {|r|r|r|r|r|r|r|r|r|} \hline P &Q &R &\neg Q &P\rightarrow Q &R\vee \neg Q & \neg Q \wedge (P\rightarrow Q) \wedge (R\vee \neg Q) \\ \hline T &T &T &F &T &T &F \\ \hline T &T &F &F &T &F &F \\ \hline T &F &T &T &F &T &F \\ \hline T &F &F &T &F &T &F \\ \hline F &T &T &F &T &T &F \\ \hline F &T &F &F &T &F &F \\ \hline F &F &T &T &T &T &T \\ \hline F &F &F &T &T &T &T \\ \hline \end{array}

Note that, in the last column, only the last two entries are true. By looking at the corresponding entries of P, Q, and R, it does not matter what value R is taking (P and Q should both take value false). Therefore, the simplified value is $\neg$ P $\wedge \neg$ Q.

Bableen Kaur
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~Q ∧ (P→Q) ∧ (R ∨ ~Q)
   = ~Q ∧ (~P ∨ Q) ∧ (R ∨ ~Q)
   = ((~Q ∧ ~P) ∨ (~Q ∧ Q)) ∧ (R ∨ ~Q)
   = ((~Q ∧ ~P) ∨ FALSE) ∧ (R ∨ ~Q)
   = (~Q ∧ ~P) ∧ (R ∨ ~Q)
   = (~Q ∧ ~P ∧ R) ∨ (~Q ∧ ~P ∧ ~Q)
   = (~Q ∧ ~P ∧ R) ∨ (~Q ∧ ~P)
   = (~Q ∧ ~P ∧ R) ∨ (~Q ∧ ~P ∧ TRUE)
   = (~Q ∧ ~P) ∧ (R ∨ TRUE)
   = (~Q ∧ ~P) ∧ TRUE
   = (~Q ∧ ~P)
CiaPan
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