The statement \[(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))\] is equivalent to \_\_\_\_\_\_\_\_\_.
Solutions · 1
Official solution
\[
\[pTTFFqTFTF∼qFTFT∼pFFTTp∧∼qFTFF∼p∧qFFTF∼p∧∼qFFFT\]
\]
Using the truth table, we can see that the given statement is equivalent to \[(\sim p) \vee(\sim q)\].
Alternate Method :
The given statement \[(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))\] can be simplified using the following steps:
\[
(p∧(∼q))∨((∼p)∧q)∨((∼p)∧(∼q))=(p∧(∼q))∨((∼p)∧(q∨(∼q)))=(p∧(∼q))∨((∼p)∧t)=(p∧(∼q))∨(∼p)=(∼p)∨(p∧(∼q))=(∼p∨p)∧(∼p∨∼q)=t∧(∼p∨∼q)=∼p∨∼q
\]
Therefore, the given statement is equivalent to \[(\sim p) \vee (\sim q)\].