Question:

Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer: Rewrite it in precise format : Problem (a) **Premises:** P1: $(p\land q)\lor(r\land s)$ P2: $p \Rightarrow \lnot p$ **Goal:** derive $r$. Formal proof: 1. $(p\land q)\lor(r\land s)$. — Premise (P1) 2. $p \Rightarrow \lnot p$. — Premise (P2) 3. $\lnot p \lor \lnot p$. — from (2) by **E18** (implication → disjunction): $α⇒β \equiv \lnot α\lor β$. * (apply with $α=p,\ β=\lnot p$). 4. $\lnot p$. — from (3) by **E4** (idempotent law) since $\lnot p\lor\lnot p \equiv \lnot p$. 5. $\lnot p \lor \lnot q$. — from (4) by **I1** (addition): from $\lnot p$ infer $\lnot p\lor\lnot q$. 6. $\lnot(p\land q)$. — from (5) by **E16** (DeMorgan: $\lnot(\alpha\land\beta)\equiv\lnot\alpha\lor\lnot\beta$), read backwards. * (So we have now $\lnot(p\land q)$.) 7. $ (p\land q)\lor(r\land s)$. — repeat of (1) for clarity (or use commutativity E9 to reorder if you prefer). 8. $r\land s$. — from (7) and (6) by **Disjunctive Syllogism (DS)**: $(A\lor B),\ \lnot A \;\vdash\; B$. * Here $A$ is $(p\land q)$ and $B$ is $(r\land s)$. Since we have $\lnot(p\land q)$ (line 6), DS yields $r\land s$. 9. $r$. — from (8) by **I2** (simplification): from $r\land s$ infer $r$. **Conclusion:** $r$ (line 9). □ --- ## Problem (j) **Premises:** P1: $\lnot q \lor p$ P2: $\lnot r \lor (s \land \lnot p)$ P3: $\lnot t \lor r \lor q$ **Goal:** show $\lnot s \Rightarrow (\lnot t \lor p)$. Equivalently (using implication ↔ disjunction): show $s \lor \lnot t \lor p$. I will do a conditional proof: assume $\lnot s$ and derive $\lnot t \lor p$. Formal proof (conditional style): 1. $\lnot q \lor p$. — Premise (P1) 2. $\lnot r \lor (s \land \lnot p)$. — Premise (P2) 3. $\lnot t \lor r \lor q$. — Premise (P3) 4. Assume $\lnot s$. — Assumption for conditional proof (we will derive $\lnot t \lor p$ under this assumption). 5. From (2) and (4) infer $\lnot r$. — reasoning: with $\lnot s$ the disjunct $s\land\lnot p$ is false, so the only way $\lnot r \lor (s\land\lnot p)$ can be true is $\lnot r$. Formally we can proceed as follows: * 5a. From (4) by **I1** (addition) get $\lnot s \lor \lnot p$. (Optional intermediate.) * 5b. Note $(s\land\lnot p) \Rightarrow s$. From $\lnot s$ we get $\lnot(s\land\lnot p)$ (DeMorgan and addition, similar to technique in (a)). * 5c. Since $\lnot(s\land\lnot p)$, and we have $\lnot r \lor (s\land\lnot p)$, apply DS to conclude $\lnot r$. (So final compact justification: from (2) and $\lnot(s\land\lnot p)$ apply DS → $\lnot r$.) I will cite the result as: $\lnot r$. — from (2),(4) by DeMorgan + DS (as in (a)). 6. From (3) and (5) infer $\lnot t \lor q$. — from $\lnot t \lor r \lor q$ and $\lnot r$ remove $r$ by DS on the subdisjunction: more concretely: rearrange (3) (commutativity) to $(r \lor q) \lor \lnot t$, use DS with $\lnot r$ to get $\lnot t \lor q$. So: $\lnot t \lor q$. 7. Now consider $\lnot t \lor q$ (line 6) and $\lnot q \lor p$ (line 1). We want $\lnot t \lor p$. Use cases: * From $\lnot t \lor q$: if $\lnot t$ holds, we are done (the disjunction $\lnot t \lor p$ is true). * If $q$ holds, then from (1) ($\lnot q \lor p$) and $q$ we can infer $p$. Formally: $q$ and $\lnot q \lor p$ give $p$ by DS/rearrangement and modus ponens style reasoning (or think of P1 as $q\Rightarrow p$, then apply modus ponens). More concretely: * 7a. $\lnot q \lor p$ (line 1) is equivalent by E18 to $q \Rightarrow p$. * 7b. If we have $q$, then by modus ponens (I3) we get $p$. * Thus from $\lnot t \lor q$ and (1) we can conclude $\lnot t \lor p$: either $\lnot t$ holds or (if $q$) we get $p$. Formally combine these as: from (6) and (1) derive $\lnot t \lor p$. (This is a valid, standard inference: take $\lnot t \lor q$ and replace the $q$-branch with $p$ using $q\Rightarrow p$.) 8. Therefore under assumption $\lnot s$ we have derived $\lnot t \lor p$. So discharge the assumption and conclude: $$ \lnot s \Rightarrow (\lnot t \lor p). $$ — By conditional proof (assumption discharged). **Conclusion:** $\lnot s \Rightarrow (\lnot t \lor p)$. □