Suppose \(A\) and \(B\) are false statements. Is the statement \[(A \implies B) \implies (A \vee B)\] true or false? (“\(\vee\)” denotes “or.”)
Give the negation of the statement
If all blockoids are split and some blockoid is nontrivial, then there is a short blockoid.
Solution
Part (a).
1. \(A \implies B\) is true.
\(A\) is false, and a false antecedent makes an implication true.
2. \(A \vee B\) is false.
both \(A\) and \(B\) are false.
3. Hence \((A \implies B) \implies (A \vee B)\) is \(\text{true} \implies \text{false}\), which is false.
an implication with a true antecedent and false consequent is false.
4. Q.E.D.
3.
Part (b).
1. The statement is of the form \(P \implies Q\), where \(P = (\text{all blockoids are split}) \wedge (\text{some blockoid is nontrivial})\) and \(Q = (\text{there is a short blockoid})\).
parse the statement.
2. The negation of \(P \implies Q\) is \(P \wedge \neg Q\).
\(\neg(P \implies Q) \equiv P \wedge \neg Q\).
3. Hence the negation is: “All blockoids are split, and some blockoid is nontrivial, and there is no short blockoid.”
1 and 2.
4. Q.E.D.
3.