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+17 votes
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Choose the correct alternatives (more than one may be correct) and write the corresponding letters only:

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.
asked in Mathematical Logic by Veteran (59.5k points)
edited by | 914 views
–1
Plz someone explain why a,c,d are wrong?

4 Answers

+18 votes
Best answer

answer = option (B)

$$\text{False} \rightarrow \text{anything} = \text{True, always}$$

answered by Boss (30.7k points)
edited by
0
How B is right answer?
+2

vaishali not satisfiable means contadiction only[ false = .F1∧F2]

we know that False→anything=True, always

+1
The conjunction F1∧F2 is not satisfiable:
meaning of this statement is , F1∧F2 is always false.

Am i right?
+2
yes .. then B is correct na?
0
Yes..my mistake.
+2
.....:) Do all mistake before gate :)
0
haha...yes u r right!!
0
How is RHS is always true?

Can i re-write as:-

f1 ^  f2 ==> f3 ^ ~f3,here it will become f1 ^  f2 ==>0,

or

f1 ^  f2 ==> f3 or ~f3,here it will become f1 ^  f2 ==>1,means lhs should be 0 if rhs is always true.

Which version is correct?Or is their some other approach we follow?If this this wrong approach then please tell why?
+1
It is easy @rahul. F1 ^ F2 --> F3 and F1 ^ F2 ---> ~F3. Both these are tautologies.
Can you consider F1 ^ F2 as true and prove that both these are tautologies? Try it.
+3
Got it.If i make truth table ,then i can see that when conjunction is true then the result of the implication in dependent on f3.So in case conjunction is true only one of them becomes true.But as we know both are tautologies so we take conjunction as false.Is this correct resoning?
0
How ?
+3 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.

answered by Active (4k points)
+2 votes
" F1∧F2→F3 and F1∧F2→∼F3 are both tautologies " it is possible in 2 cases
case 1) True→True
case 2) False→False/True
here F3 is in both F3 and ∼F3 form so only case 2) will apply
so F1∧F2 is False means F1=False and F2=False

(a). "Both F1 and F2 are tautologies" is INCORRECT

(b). "The conjunction F1∧F2 is not satisfiable" is INCORRECT becoz for being satisfiable atleast one possibility of F1 and F2 should be True

(c). "Neither is tautologous" is INCORRECT as both are Tautologies

(d). "Neither is satisfiable " is INCOORECT as both are Tautologies so also satisfiable

Hence correct ans is B
answered by Active (2.3k points)
edited by
0
How can we prove that these two are tautologies [email protected] Ranker18 explain this pls
0
@set2018

which two ?
0
F1∧F2→F3 and F1∧F2→∼F3 are both tautologies..................How Pls explain this .
0
it is given in ques
0 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
answered by Loyal (6.5k points)
0
in option d) when no body is satisfiable.

it means,   F->F

                 F->T

both are always true .

so it will be tautology


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