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Show that ∼(p∨(∼p∧q))≡∼p∧∼q (By using Laws of Logic)
Starting Date: May 11, 2016.
Due Date: May 16, 2016.
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MTH202 - Discrete Mathematics GDB No. 1 Solution Spring 2016 Due Date May 16, 2016
Starting Date: May 11, 2016.
Due Date: May 16, 2016.
Dear fellows!
It is intimated that this will not be solved with truth table!
It will be solved using laws of logic that are kind of some properties etc etc
lawas of logic use hon gy iss mein like distibutive assosiative law etc
first use Distributive law.
Then use demorgan's law
then again use demorgan's law
then apply double negation law
then apply idempotent law
then again apply indempotent law.
....
Please solve it yourself. I have attached laws which are goin to be used.
MTH202 gdb Solution may 2016
MTH-202 GDB-1
Complete Solution
Show that ∼(p ∨ (∼p ∧ q)) ≡ ∼p ∧ ∼q (By using Laws of Logic)
Solution:
L.H.S
∼(p ∨ (∼p ∧ q)) ≡ ~p ∧ ~(∼p ∧ q) Demorgans Law
≡ ~p ∧ (∼(∼p) ∨ ∼ q)
≡ ~p ∧ (p ∨ ∼ q) Double negation law
≡ (~p ∧ p) ∨ (~p ∧ ∼ q) 2^{nd} Distributive law
≡ F ∨ (~p ∧ ∼ q) As ~p ∧ p=F
≡~p ∧ ∼ q Identity law
Hence Proved
sis it true ? which one solution i
~(p∨(∼p∧q))≅∼p∧∼q
L.H.S
≅~[p∨(∼p∧q)] by Applying distributive law [p ∨ ( q ∧ r ) ≡ ( p ∨ q ) ∧ ( p ∨ r )]
≅~[(p∨∼p)∧ (p∨q)]
≅~[t ∧ (p∨q)] by Applying negation law [p ∨ ∼p ≡ t]
≅~(p∨q) by applying identity law [ p ∧ t ≡ p ]
≅~p∧~q by applying DE Morgan’s law [ ~ ( p ∨ q ) ≡ ~p ∧ ∼q]
Hence proved L.H.S ≅ R.H.S
~p∧~q≅~p∧~q
∼(p∨(∼p∧q))≡∼p∧∼q
Solution:-
≡~(p∨(∼p∧q))
≡~((p∨~p)∧(p∨q)) Reverse Distributive Law
≡~(p∨~p)∨~(p∨q) Demorgan Law
≡(~(p)∧~(~p))∨(~(p)∧~(q)) Demorgan Law
≡(~p∧p)∨(~p∧~q) Negation Law
≡~p∨(~p∧~q) Negation Law
≡(~p∨~p)∧(~q) Associative Law
≡~p∧~q Idempotent Law
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