MTH202 Assignment no 01 2016 - Solution


MTH202 Assignment no. 01

MTH202 Discrete Mathematics Assignment no 01 fall 2016 has been uploaded. You can download the Assignment and Solution File from the Link below.The last date of submission is November 14, 2016.

Assignment No. 1 MTH202 (Fall 2016)
Maximum Marks: 10 
Due Date: 14 -11-2016 

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MTH202 Assignment:

Download Complete Assignment: Fall 2016_MTH202_Assignment_1.doc, 30 KB

MTH202 Assignment Solution:

MTH202 Assignment no 01 Solution has been uploaded.
Please Try to make you own if you feel something is wrong correct me with valid logic because this is logical assignment.
2nd question I solved with different method. 
Take Help from my solution apply laws of logic.
  1. Commutative laws: p∧q ≡ q∧p p∨q ≡ q∨p
  2. Associative laws: (p∧q)∧r ≡ p∧(q∧r) (p∨q)∨r ≡ p∨(q∨r)
  3. Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r) p∨(q∧r) ≡ (p∨q)∧(p∨r)
  4. Identity laws: p∧t ≡ p p∨c ≡ p
  5. Negation laws: p∨∼p ≡ t p∧∼p ≡ c
  6. Double negative law: ∼(∼p) ≡ p
  7. Idempotent laws: p∧p ≡ p p∨p ≡ p
  8. Universal bound laws: p∨t≡t p∧c≡c
  9. De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ∼(p∨q) ≡ ∼p∧∼q
  10. Absorption laws: p∨(p∧q) ≡ p p∧(p∨q) ≡ p
  11. Negations of t and c: ∼t ≡ c ∼c ≡ t
The first circuit is equivalent to this: (P∧Q) ∨ (P∧~Q) ∨ (~P∧~Q), which I managed to simplify to this: P ∨ (~P∧~Q).

The other circuit is simply this: P ∨ ~Q

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Download Assignment Solution: MTH202 Solution.doc, 89 KB

1 comment:

  1. Our main purpose here discussion not just Solution