Suppose the propositional variable p has truth value 1. Determine all truth values of q, r and s for which (p → ((¬p ∨ r) ∧ ¬s)) ∧ (¬s → (¬r ∧ p)) has truth value 1.

Step-by-step solution

1. 1

   Notice that q never appears. The variable $q$ occurs nowhere in the formula, so its value is irrelevant — whatever conclusion is reached will hold for both $q=0$ and $q=1$. Spotting this first removes half the case work.

2. 2

   Reduce the first conjunct using p = 1. With $p=1$ we have $\neg p=0$, so $\neg p\vee r=r$. And since $1\to X\equiv X$,

   $p\to[(\neg p\vee r)\wedge\neg s]\;\equiv\;r\wedge\neg s$

   For the whole conjunction to be $1$, this forces $r=1$ and $s=0$.

3. 3

   Reduce the second conjunct using p = 1. Here $\neg r\wedge p=\neg r$, so

   $\neg s\to(\neg r\wedge p)\;\equiv\;\neg s\to\neg r\;\equiv\;s\vee\neg r$

4. 4

   Combine the two demands. The formula is $1$ exactly when

   $(r\wedge\neg s)\wedge(s\vee\neg r)=1$

   The left factor requires $r=1,\ s=0$; substituting those into the right factor gives $s\vee\neg r=0\vee 0=0$. The conjunction is therefore $0$.

5. 5

   Conclude that the formula is unsatisfiable at p = 1.

   $\boxed{\text{No values of }q,r,s\text{ make the formula true when }p=1}$

   On a multiple-choice version listing $(r,s)=(0,1),(1,1),(1,0),(0,0)$, every option is wrong.

6. 6

   Verify by exhaustive truth table. With $p=1$, check the four $(r,s)$ pairs against $(r\wedge\neg s)\wedge(s\vee\neg r)$: $(0,0)\to 0\wedge 1=0$; $(0,1)\to 0\wedge 1=0$; $(1,0)\to 1\wedge 0=0$; $(1,1)\to 0\wedge 1=0$. All four give $0$, confirming the algebraic reduction. Equivalently, $r\wedge\neg s$ and $s\vee\neg r=\neg(r\wedge\neg s)$ are exact negations of each other, so their conjunction is a contradiction for any $p$.