Question 2 of 12

Question 2 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) Two logical forms in $p,q,r$. (b) A conditional whose antecedent is the fixed proposition "$2\gt 10$". (c) The existential statement from Q1(b). (d) The proposition $(\neg p\land q)\lor\neg r$.

Find. (a) A proof of the equivalence (truth table or logical laws). (b) The truth value. (c) The negation, simplified. (d) The dual (swap $\land\leftrightarrow\lor$; no constants to swap here).

Approach. (a) build the 8-row truth table for $p,q,r$ (or push the negation using $p\to r\equiv\neg p\lor r$); (b) a conditional with a false antecedent is vacuously true, independent of the consequent; (c) push $\neg$ through the quantifier and the inequality; (d) apply the duality principle.

1. (a) Prove $(p\to r)\lor(q\to r)\equiv(p\land q)\to r$. Rewrite every conditional with $\to$ eliminated: $p\to r\equiv\neg p\lor r$, so the left side is $\neg p\lor r\lor\neg q\lor r \equiv \neg p\lor\neg q\lor r$. The right side is $(p\land q)\to r\equiv \neg(p\land q)\lor r \equiv (\neg p\lor\neg q)\lor r$ by De Morgan. Both sides reduce to the identical clause $\neg p\lor\neg q\lor r$: $$(p\to r)\lor(q\to r)\ \equiv\ \neg p\lor\neg q\lor r\ \equiv\ (p\land q)\to r$$ A full 8-row truth table over $p,q,r$ confirms both sides agree on every row. $\boxed{(p\to r)\lor(q\to r)\equiv(p\land q)\to r}$
2. (b) Truth value of "If $2\gt 10$ then $\forall x\,x=x+1$". The antecedent "$2\gt 10$" is false. A conditional with a false antecedent is true regardless of the consequent (vacuous truth) — it does not matter that "$\forall x\,x=x+1$" is itself false. $\boxed{\text{TRUE (vacuously)}}$
3. (c) Negation of $\exists n\,(n+1\gt n^2)$. Push the negation through the existential quantifier (turning it universal) and negate the inequality: $$\neg\exists n\,(n+1\gt n^2)\ \equiv\ \forall n\,\neg(n+1\gt n^2)\ \equiv\ \forall n\,(n+1\le n^2)$$ $\boxed{\forall n\,(n+1\le n^2)}$
4. (d) Dual of $(\neg p\land q)\lor\neg r$. The dual of a proposition built only from $\land,\lor$ (no $T/F$ constants) swaps every $\land\leftrightarrow\lor$ and leaves the literals untouched: $$(\neg p\land q)\lor\neg r\quad\xrightarrow{\text{dual}}\quad(\neg p\lor q)\land\neg r$$ $\boxed{(\neg p\lor q)\land\neg r}$