The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
\((p∧(∼r))⇒q\)
\((∼q)⇒((∼r)∨p)\)
\(p⇒(q∨r)\)
\((p∧(∼q))⇒r\)
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:
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:
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: \((\sim q) \Rightarrow ((\sim r) \lor p)\)
\((A)(p∧(∼r))⇒q\)
\(∼(p∧∼r)∨q\)
\(≡(∼p∨r)∨q\)
\(≡∼p∨(r∨q)\)
\(≡p→(q∨r)\)
\(≡(p⇒q)∨(p⇒r)\)
\((C)p⇒(q∨r)\)
\(≡∼p∨(q∨r)\)
\(≡(∼p∨q)∨(∼p∨r)\)
\(≡(p→q)∨(p→r)\)
\((D)(p∧∼q)⇒r\)
\(≡p⇒(q∨r)\)
\(≡(p⇒q)∨(p⇒r)\)
So, the correct option is (B): \((∼q)⇒((∼r)∨p)\)
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