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40 votes
40 votes

If $F_1$, $F_2$ and $F_3$ are propositional formulae such that $F_1 \land F_2 \rightarrow F_3$ and $F_1 \land F_2 \rightarrow \sim  F_3$ are both tautologies, then which of the following is true:

  1. Both $F_1$ and $F_2$ are tautologies
  2. The conjunction $F_1 \land F_2$ is not satisfiable
  3. Neither is tautologous
  4. Neither is satisfiable
  5. None of the above
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9 Answers

7 votes
7 votes

F1∧F2→F3

F1∧F2→∼F3 

Now take contrapositive of both

∼F3→∼(F1∧F2)

  F3→∼(F1∧F2)

--------------------------------------------by using resolution rule

∼(F1∧F2) or ∼(F1∧F2)=∼(F1∧F2)

which means Nand of f1 and f2 is tautology ,therefore conjuction of both of them is contradiction hence not satisfiable.

so option B is th answer.

5 votes
5 votes
Solved by Argument method ,

Let P1: F1∧F2→F3 and P2: F1∧F2→∼F3 is two premises ,so we need to findout what is the conclusion .

P1: F1∧F2→F3

P2: F1∧F2→∼F3

Both the premises will simultaneously true when (F1∧F2) is False i.e ~(F1∧F2).

Hence the conclusion is ~(F1∧F2) i.e the conjunction (F1∧F2)is not satisfiable.

So ans is (b). The conjunction F1∧F2 is not satisfiable
1 votes
1 votes

Given, $F_1 \wedge F_2 \implies F_3$ is $True$. 

$F_1 \wedge F_2 \implies  \neg F_3$ is $True$.

Notice that, the antecedent is same for both the statements but the conclusion is different. Whenever $F_3$ is $True$, then $\neg F_3$ is $False$ and vice-versa. So, for both the statements to be tautology the antecedent is the defining proposition. So, antecedent needs to be $False$, so as to predict $True$ for both the statements. Thus, $F_1 \wedge F_2$ is $False$, i.e, $Not$ $Satisfiable$.

Thus, option B is correct.

0 votes
0 votes
precedence of and operator is higher than implies operator .

 

F1^F2->F3 =F1.F2->F3 -(1)

 

F1^F2->~F3 = F1.F2->~F3 (2)

 

FROM eq. (1) & (2)

 

F1.F2->0 IS TRUE ONLY IF F1.F2 false

 

SO THAT IT NEVER BE Satisfiable.

 

 

option B is correct option
Answer:

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