Question

The statement 
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to

• A. \((p∧(∼r))⇒q\)
• B. \((∼q)⇒((∼r)∨p)\)
• C. \(p⇒(q∨r)\)
• D. \((p∧(∼q))⇒r\)

Answer 1

The Correct Option is B

Let's analyze the given logical statement and its options to find which option is NOT equivalent to the statement \((p \Rightarrow q) \lor (p \Rightarrow r)\). 

First, we will recall some fundamental concepts:

• The implication \(p \Rightarrow q\) is equivalent to \((\sim p) \lor q\).
• The equivalent transformation of multiple logical expressions helps in determining their equivalence.

Now, let's transform the given statement:

\((p \Rightarrow q) \lor (p \Rightarrow r)\) is equivalent to \(((\sim p) \lor q) \lor ((\sim p) \lor r)\).

Using the distributive law:
\(((\sim p) \lor q) \lor ((\sim p) \lor r) = (\sim p) \lor (q \lor r)\)

This further simplifies to:
\(p \Rightarrow (q \lor r)\)

Now, let's compare each option with the simplified statement:

1. \((p \land (\sim r)) \Rightarrow q\)
   This statement implies that if both \(p\) and \(\sim r\) are true, then \(q\) must be true. It is not directly equivalent to the original statement.
2. \((\sim q) \Rightarrow ((\sim r) \lor p)\)
   Transforming, \(\sim q \Rightarrow (\sim r \lor p)\) becomes \(q \lor (\sim r \lor p)\), which is not equivalent to the simplified form.
3. \(p \Rightarrow (q \lor r)\)
   Already derived as the simplified version of the original statement.
4. \((p \land (\sim q)) \Rightarrow r\)
   This implies that if both \(p\) and \(\sim q\) are true, then \(r\) must be true. It could be considered equivalent in specific cases but doesn't match the general form derived above.

From the above analysis, it is clear that the option \((\sim q) \Rightarrow ((\sim r) \lor p)\) does not match the equivalent transformation of the statement \((p \Rightarrow q) \lor (p \Rightarrow r)\).

Therefore, the correct answer is: \((\sim q) \Rightarrow ((\sim r) \lor p)\)

Answer 2

\((A)(p∧(∼r))⇒q\)
\(∼(p∧∼r)∨q\)
\(≡(∼p∨r)∨q\)
\(≡∼p∨(r∨q)\)
\(≡p→(q∨r)\)
\(≡(p⇒q)∨(p⇒r)\)
\((C)p⇒(q∨r)\)
\(≡∼p∨(q∨r)\)
\(≡(∼p∨q)∨(∼p∨r)\)
\(≡(p→q)∨(p→r)\)
\((D)(p∧∼q)⇒r\)
\(≡p⇒(q∨r)\)
\(≡(p⇒q)∨(p⇒r)\)
So, the correct option is (B): \((∼q)⇒((∼r)∨p)\)