The boolean [removed]¬p→(p∧q))∨q is equivalent to:
The boolean [removed]¬p→(p∧q))∨q is equivalent to:
Option 1 -
q→(p∧q)
Option 2 -
p→q
Option 3 -
p→(p→q)
Option 4 -
p→(p∨q)
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1 Answer
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Correct Option - 4
Detailed Solution:| p | q | p^q | ¬ (p^q) | (¬ (p^q)vq | pvq | p? q | p? (pvq) |
|-|-|-|-|-|-|-|-|
| T | T | T | F | T | T | T | T |
| T | F | F | T | T | T | F | T |
| F | T | F | T | T | T | T | T |
| F | F | F | T | T | F | T | T |
Tautology, Tautology
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Case – I
it can be false if r is false,
so not a tautology
Case – II If
tautology
then
Case – I
pva
its negation as asked in question
=
=
mod (7)
…… (i)
Now,
……. (ii)
(i) & (ii)
kindly consider the following Image

q is equivalent to
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