Which of the following is true ? (a) \(\mathrm{p} \Lambda(\sim \mathrm{p})=\mathrm{t}\) (b) \(p V(\sim p)=f\) (c) \(p \Rightarrow q=q \Rightarrow p\) (d) \(\mathrm{p} \Rightarrow \mathrm{q}=(\sim \mathrm{q}) \Rightarrow(\sim \mathrm{p})\)

We are given that \(\mathrm{p} \Lambda(\sim \mathrm{p})=\mathrm{t}\), and we will need to check if it's true. \(\mathrm{p} \Lambda(\sim \mathrm{p})\) means \(\mathrm{p}\) AND NOT \(\mathrm{p}\). The truth table for AND is as follows: | p | NOT p | p AND NOT p | |---|-------|-------------| | T | F | F | | F | T | F | As we can see, both combinations of p (T or F) give us F for the compound statement \(\mathrm{p} \Lambda(\sim \mathrm{p})\). Thus, the statement (a) is false, as it is not equal to t.

We are given that \(p V(\sim p)=f\), and we will need to check if it's true. \(p V(\sim p)\) means \(p\) OR NOT \(p\). The truth table for OR is as follows: | p | NOT p | p OR NOT p | |---|-------|------------| | T | F | T | | F | T | T | As we can see, both combinations of p (T or F) give us T for the compound statement \(p V(\sim p)\). Thus, the statement (b) is false, as it is not equal to f.

We are given that \(p \Rightarrow q=q \Rightarrow p\), and we need to check if they are equivalent. \(p \Rightarrow q\) means IF \(p\), THEN \(q\). The truth table for IMPLIES is as follows: | p | q | p IMPLIES q | q IMPLIES p | |---|---|-------------|-------------| | T | T | T | T | | T | F | F | T | | F | T | T | F | | F | F | T | T | As we can see, in two rows (second row and third row) the results for \(p \Rightarrow q\) and \(q \Rightarrow p\) are different. Thus, the statement (c) is false, as \(p \Rightarrow q\) is not equivalent to \(q \Rightarrow p\).

We are given that $\mathrm{p} \Rightarrow \mathrm{q}=(\sim \mathrm{q}) \Rightarrow(\sim \mathrm{p})$, and we will need to check if they are equivalent. The truth table is as follows: | p | q | NOT q | NOT p | p IMPLIES q | NOT q IMPLIES NOT p | |---|---|-------|-------|-------------|---------------------| | T | T | F | F | T | T | | T | F | T | F | F | F | | F | T | F | T | T | T | | F | F | T | T | T | T | As we can see, for all combinations of p and q, the results for \(\mathrm{p} \Rightarrow \mathrm{q}\) and \((\sim \mathrm{q}) \Rightarrow(\sim \mathrm{p})\) are equal. Thus, statement (d) is true. The correct answer is: (d) \(\mathrm{p} \Rightarrow \mathrm{q}=(\sim \mathrm{q}) \Rightarrow(\sim \mathrm{p})\)