Logic

“Exactly one mother” means: \(x\) has at least one mother \(y\), and nobody other than \(y\) is a mother of \(x\).

• (A) False: it only says \(x\) has some mother \(y\) and some non-mother \(z\); a second mother is not ruled out.
• (B) True: \(y\) is a mother of \(x\), and every \(z\) different from \(y\) is not a mother of \(x\).
• (C) False: it says that if \(y\) is a mother of \(x\), then some \(z\) equal to \(y\) is a mother of \(x\). That is always true (take \(z = y\)), so it neither forces a mother to exist nor limits how many there are.
• (D) True: \(\lnot \exists z\, (noteq(z, y) \land mother(z, x))\) is equivalent to \(\forall z\, (noteq(z, y) \to \lnot mother(z, x))\), so (D) says the same as (B).

Answer: B, D

Option A is the principle of mathematical induction: \(P(0)\) holds, and truth passes from each \(x\) to \(x + 1\), so it reaches 1, 2, 3, … and hence every natural number. It is true for every predicate \(P\).

Each other option fails for some predicate:

• B: take \(P(x)\) to be “\(x \le 5\)”. Both premises hold, but \(P(6)\) is false.
• C: take “\(x \le 1000\)”. Both premises hold, but \(P(1001)\) is false.
• D: take “\(x \ge 1000\)”. Both premises hold, but \(P(0)\) is false.

Answer: A

“\(X\) when \(Y\)” means “whenever \(Y\) holds, \(X\) holds”, which is the implication \(Y \rightarrow X\).

• The condition \(Y\) is “student scores more than 50% marks”, that is \(q\).
• The consequence \(X\) is “fail grade cannot be given”, that is \(\neg p\).

So the statement is \(q \rightarrow \neg p\). Option B drops the negation and says a fail grade can be given. Option D, \(\neg p \rightarrow q\), is the converse of \(q \rightarrow \neg p\) and is not equivalent to it, and option C has neither the right direction nor the negation.

Answer: A