GO Classes CS 2025 | Weekly Quiz 1 | Propositional Logic | Question: 13

Answer 1

Answer: 3

Let’s simplify the propositional function a bit:

$= \lnot p \leftrightarrow (q \land \lnot(p \to r))$

$= \lnot p \leftrightarrow (q \land \lnot (\lnot p \lor r))$

$= \lnot p \leftrightarrow (q \land p \land \lnot r)$

Now we know $a \leftrightarrow b$ is true when either both $a$ and $b$ are false or both $a$ and $b$ are true.

Now considering case (i) when both $a$ and $b$ are false 

It is only possible when $p$ is true so $\lnot p$  is false, now to $(q \land p \land \lnot r)$ to be false we have following options 

 

$q$	$r$	$(q∧p∧¬r)$
F	F	F
F	T	F
T	F	T
T	T	F

So $(q∧p∧¬r)$ is false only for 3 options.

 

Now considering case (ii) when both $a$ and $b$ are true 

Here $a$ to be true $p$ must be false but if $p$ is false $(q∧p∧¬r)$ can’t be true.

So total combination of $p$, $q$ and $r$ for which $\lnot p \leftrightarrow (q \land \lnot(p \to r))$ is true is only 3.

Answer 2

only 3 comb is possible 

Answer 3

case1 : p=false

p’ $\leftrightarrow$ ( q (p → r)’ ) $\equiv$ T  => T$\leftrightarrow$F $\equiv$ T  => F $\equiv$ T. Contradiction. p=false is not possible.

case2 : p=true

p’ $\leftrightarrow$ ( q (p → r)’ ) $\equiv$ T  => F$\leftrightarrow$ (qr’) $\equiv$ T  => (qr’) $\equiv$ F  => 3 possible combinations of q & r.

Ans is 3. (p is true and 3 possible combinations of q & r such that qr’ is false)

Answer 4

Thanks you!!