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Assignment 2 2. Which of the following formulas are tautologies? Justify your answers. (a) (P+ (PVO))
Assignment 2 (b) ((PA-Q) + R)
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Answer #1

If a statement is a tautology, then all the values in the final column of its truth table will be true. i.e. The statement will be true for all the cases.

So construct a truth table for each of the statements to check whether they are tautology.

a. Consider (P\rightarrow \left ( P\vee Q \right ))

Now Construct a truth table for the formula (P\rightarrow \left ( P\vee Q \right )) :

P Q \left ( P\vee Q \right ) (P\rightarrow \left ( P\vee Q \right ))
0 0 0 1
0 1 1 1
1 0 1 1
1 1 1 1

Notice that the given formula (P\rightarrow \left ( P\vee Q \right )) is true for all the possible cases.

hence (P\rightarrow \left ( P\vee Q \right )) is a tautology.

.

(b). Consider \left (\left (P\wedge \sim Q \right )\rightarrow R \right ) .

Now Construct a truth table for the formula \left (\left (P\wedge \sim Q \right )\rightarrow R \right ) :

P Q R \sim Q \left (P\wedge \sim Q \right ) \left (P\wedge \sim Q \right )\rightarrow R
0 0 0 1 0 1
0 0 1 1 0 1
0 1 0 0 0 1
0 1 1 0 0 1
1 0 0 1 1 0
1 0 1 1 1 1
1 1 0 0 0 1
1 1 1 0 0 1

From the truth table, note that the formula \left (\left (P\wedge \sim Q \right )\rightarrow R \right ) is not true for all cases. It is false when P is true and Q and R are both false. Hence the given formula \left (\left (P\wedge \sim Q \right )\rightarrow R \right ) is not a tautology.

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