Question

The negation of the statement \(((A\land(B\lor C)) ⇒ (A\lor B)) ⇒ A\) is

• A. a fallacy
• B. equivalent to \(B\lor \sim C\)
• C. equivalent to \(\sim A\)
• D. equivalent to \(\sim C\)

Answer 1

The Correct Option is C

Let us denote \[ p := \Bigl(\bigl(A \wedge (B \vee C)\bigr) \Rightarrow (A \vee B)\Bigr) \Rightarrow A. \] First, recall that an implication \(X \Rightarrow Y\) is logically equivalent to \(\lnot X \vee Y\). Thus \[ (A \wedge (B \vee C)) \Rightarrow (A \vee B) \longleftrightarrow \lnot\bigl(A \wedge (B \vee C)\bigr) \vee (A \vee B). \] Hence \[ p \longleftrightarrow \Bigl(\lnot\bigl(A \wedge (B \vee C)\bigr) \vee (A \vee B)\Bigr) \Rightarrow A. \] Again using \(X \Rightarrow A\) is \(\lnot X \vee A\), we get \[ p \longleftrightarrow \lnot\Bigl(\lnot\bigl(A \wedge (B \vee C)\bigr) \vee (A \vee B)\Bigr) \vee A. \] Distributing the negation inside, \[ \lnot\Bigl(\lnot\bigl(A \wedge (B \vee C)\bigr) \vee (A \vee B)\Bigr) \longleftrightarrow \bigl(A \wedge (B \vee C)\bigr) \wedge \lnot(A \vee B). \] Therefore \[ p \longleftrightarrow \Bigl[\bigl(A \wedge (B \vee C)\bigr) \wedge \lnot(A \vee B)\Bigr] \vee A. \] Because the part \(\bigl(A \wedge (B \vee C)\bigr) \wedge \lnot(A \vee B)\) is inconsistent with \(A \vee B\), one can verify that this disjunction simplifies logically to just \(A\). Hence \(p \equiv A\). Thus \[ \lnot p \equiv \lnot A. \] So the negation of the original statement is logically equivalent to \(\lnot A\).