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# A GDB has been announced on VULMS Starting Date: May 11, 2016. Due Date: May 16, 2016.

Show that  ∼(p∨(∼p∧q))≡∼p∧∼q (By using Laws of Logic)

Starting Date: May 11, 2016.

Due Date:  May 16, 2016.

Note: In case you are using Math Type for mathematical expression please follow the Instructions .

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### Replies to This Discussion

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

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

Attachments:

MTH-202 GDB-1

Complete Solution

Show that  ∼(p ∨ (∼pq)) ≡ ∼p ∧ ∼q (By using Laws of Logic)

Solution:

L.H.S

∼(p ∨ (∼pq)) ≡ ~p ∧ ~(∼pq)                      Demorgans Law

≡ ~p ∧ (∼(∼p) ∨ ∼ q)

≡ ~p ∧ (p ∨ ∼ q)                     Double negation law

≡ (~p ∧ p)(~p ∧ ∼ q)          2nd 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(pq))≡p∧∼q

Solution:-

≡~(p(pq))

≡~((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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