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5 votes
5 votes

Let $A,\;B$ represent two propositions.
Which of the following logical formulae is/are tautology?

  1. $A\wedge B\rightarrow (A\rightarrow B)$
  2. $A\wedge \neg B \rightarrow \neg (A\rightarrow B)$
  3. $\neg A\wedge B \rightarrow (A\rightarrow B)$
  4. $\neg A\wedge \neg B \rightarrow (A\rightarrow B)$
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2 Answers

8 votes
8 votes
All are tautologies. These four formulas basically correspond to the truth table of implication operator.
For example, $\neg A\wedge B \rightarrow (A\rightarrow B)$ means that when $A$ is false, $B$ is true then $A \rightarrow B$ is true. Which we know is true.
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3 votes
3 votes

Let’s analyze each of the options: The approach which is used is to try and make the rhs false and lhs true or vice versa so that if this condition arrives then the given implication is false and cannot be a tautology.

  1. If we try to make A->B False then A must be true and B must be false. Now if we look at the lhs then A^B will be false . So it cannot be made true and thus the implication cannot be made false. So it is a tautology.
  2. In this case, if we try to make lhs true then A must be true and B’ must be false which means B must be true. If B is true then rhs becomes False and thus the given implication is not a tautology.
  3. In this case, if we try to make rhs false then A is True and B is false. So this means that the lhs becomes False and thus it is  a tautology.
  4. In this case, if we try to make rhs false then A is True and B is false. So this means that Lhs is False and thus it is also a tautology.
Answer:

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