Question 2 of 12

Question 2

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 universally quantified quadratic inequality over $\mathbb{Z}$, a conditional with a false antecedent, a two-clause compound proposition, and a three-variable conditional.

Find. (a), (b) truth values; (c) the simplest equivalent form; (d) a proof that the conditional is a tautology.

Approach. (a) move everything to one side and complete the square. (b) evaluate the antecedent first. (c) factor out the shared disjunct with the distributive law. (d) show the antecedent is a contradiction, so the conditional can never have a true antecedent.

1. a) $\forall n\in\mathbb{Z}\ (4n^2>4n-1)$. $4n^2-(4n-1)=4n^2-4n+1=(2n-1)^2\ge0$, with equality only at $n=\tfrac12$. Since $\tfrac12$ is not an integer, $2n-1$ is an odd integer, never $0$, so $(2n-1)^2\ge1>0$ for every integer $n$. $\boxed{\text{True}}$ (over the reals it would be false, at $x=\tfrac12$ where both sides equal 1).
2. b) "If $2>5$ then $\forall x,\ x^2\lt x^2-5$". The antecedent $2>5$ is false. (The consequent is also false, since $x^2\lt x^2-5$ reduces to $0<-5$.) A conditional with a false antecedent is true whatever the consequent: $F\to F=\boxed{\text{True}}$.
3. c) Simplify $(\neg p\lor q)\land(p\lor q)$. Both clauses share the disjunct $q$: $(q\lor\neg p)\land(q\lor p)=q\lor(\neg p\land p)=q\lor F=\boxed{q}$.
4. d) Prove $(\neg(p\lor\neg q)\land\neg q)\to r$ is a tautology. By De Morgan, $\neg(p\lor\neg q)\equiv\neg p\land q$. The antecedent is therefore $$(\neg p\land q)\land\neg q\equiv\neg p\land(q\land\neg q)\equiv\neg p\land F\equiv F.$$ The antecedent is false under every assignment, so the conditional is $F\to r\equiv T$ for every value of $r$. $\boxed{\text{Tautology}}$ (confirmed on all 8 rows of $p,q,r$).