document.write( "Question 76567: Given: (~ p → q) ∨ ~ p
\n" ); document.write( "This statement is a:
\n" ); document.write( "A. Biconditional statement B. Self-Contradiction
\n" ); document.write( "C. Tautology D. Conjunction
\n" ); document.write( "

Algebra.Com's Answer #54912 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
If we let p be false, then the entire disjunction is true since we would have\r
\n" ); document.write( "\n" ); document.write( "(~(false) → q) ∨ ~ (false)\r
\n" ); document.write( "\n" ); document.write( "(true → q) ∨ true\r
\n" ); document.write( "\n" ); document.write( "We can see that a disjunction is true if either side is true. Since the left side is true, the entire disjunction is true.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "If we let p be true, then the entire disjunction is also true since we would have\r
\n" ); document.write( "\n" ); document.write( "(~(true) → q) ∨ ~ (true)\r
\n" ); document.write( "\n" ); document.write( "(false → q) ∨ false\r
\n" ); document.write( "\n" ); document.write( "Since a false antecedent means the entire implication is true, the whole left disjunct is true. This means the whole disjunction is true. So for any p (in this case q can be anything), the logical statement is true. This means the statement is a tautology (answer c).
\n" ); document.write( "
\n" );