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Subsection 2.7.4 Contradictions

Now consider this expression:

\(p \wedge \neg p \)

Since there is only one variable, the truth table has only two rows:

p ¬p p ∧ ¬p
T F F
F T F

Now notice that the final column has all F’s. That means that, whatever the truth value for p, p ∧¬p must be false. An expression that is false for all possible truth values of its variables is called a contradiction. So we’ll say that a contradiction is unsatisfiable. Think of a contradiction as a statement that describes an impossible situation. Another way to think of it is that it describes a situation we’ll never have to worry about. We’ve now proved our first such contradiction.

Exercises Exercises

1.

Which of the following statements is a contradiction:

  1. \(\displaystyle ( p \vee q ) \rightarrow \neg p \)

  2. \(\displaystyle \neg ( p \wedge r ) \rightarrow \neg r \)

  3. \(\displaystyle ( p \vee \neg p ) \rightarrow \neg( p \vee \neg p )\)

  4. \(\displaystyle \neg q \rightarrow \neg (\neg q \wedge \neg r )\)

Answer.
Correct answer is C.
Solution.
Explanation: Write out the truth tables for all of these. When you do, you’ll see that \((p \wedge \neg p) \rightarrow \neg (p \vee \neg p) \) is the only contradiction. Of course, it’s obvious that it must be. It’s an implication that says that, if its antecedent \((p \vee \neg p) \) is true then that same expression must be false. It can’t be both true and false at the same time.

2.

Which of the following statements is a contradiction:

  1. \(\displaystyle ( q \wedge r \wedge s ) \rightarrow \neg s \)

  2. \(\displaystyle (( q \vee r ) \wedge s ) \wedge \neg ( q \wedge r \wedge s )\)

  3. \(\displaystyle ( a \wedge b ) \vee (\neg a \vee \neg b ) \)

  4. \(\displaystyle ( p \vee r) \rightarrow (p \wedge r \wedge s) \)

Answer.
Correct answer is B.
Solution.
Explanation: Write out the truth tables for all of these. When you do, you’ll see that \(((q \vee r) \wedge s) \wedge \neg(q \wedge r \wedge s) \) is the only contradiction.

Exercise Group.

Consider the following statements:

  1. \(\displaystyle ( p \wedge q ) \wedge q \)

  2. \(\displaystyle \neg ( p \wedge q ) \vee q \)

  3. \(\displaystyle ( p \rightarrow q) \wedge q \)

  4. \(\displaystyle ( p \rightarrow q) \wedge (p \wedge \neg q)\)

  5. \(\displaystyle p \wedge (q \wedge \neg r ) \)

4.

(Part 1) Which of these is a tautology?

  1. I.

  2. II.

  3. III.

  4. IV.

  5. V.

Answer.
Correct answer is B
Solution.
Explanation: Check the truth tables.
5.

(Part 2) Which of these is a contradiction (in other words, it isn’t satisfiable)?

  1. I.

  2. II.

  3. II.

  4. IV.

  5. V.

Answer.
Correct answer is D.
Solution.
Explanation: Check the truth tables.

7.

5. Using a truth table, show that the following expression is a contradiction:

\(\neg ( p \wedge ( q \vee r )) \equiv (( p \wedge q ) \vee ( p \wedge r )) \)

Click here to use the Truth Table app to do that.

www.truthtables.org/#/true/!(p&(q|r))=((p&q)|(p&r)) 2 

When you’ve finished the whole table, enter the next to last column (the one that corresponds to \(((p \wedge q) \vee (p \wedge r)) \) here. Enter a sequence of t’s and f’s, separated by a single space

Answer.
Correct answer is: t t t f f f f f
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