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Maximum Marks: 10
Due Date: 14 -11-2016
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Question No. 1:
Using the truth table show that
Question No. 2:
Using the laws of logic show that
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Discuss on Question No.2.
kon kon se laws apply hun gay is pe
aoa brothers i send u some law of Discreat Mathmatics
check these any 1,,,
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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IDEA Solution
Student Id:
MTH 202 Question No 1
p |
q |
r |
pÚq |
p→r |
q→r |
(pÚq)→r |
(p→q) Ù(q→r) |
T |
T |
T |
T |
T |
T |
T |
T |
T |
T |
F |
T |
F |
F |
F |
F |
T |
F |
T |
T |
T |
T |
T |
T |
T |
F |
F |
T |
F |
T |
F |
F |
F |
T |
T |
T |
T |
T |
T |
T |
F |
T |
F |
T |
T |
F |
F |
F |
F |
F |
T |
F |
T |
T |
T |
T |
F |
F |
F |
F |
T |
T |
T |
T |
As you show that Value of (pÚq)→r is same to (p→q) Ù(q→r) so we written.
(pÚq)→r≡(p→q) Ù(q→r)
Question No 2
p |
q |
r |
~p |
~q |
p∧r |
q∧r |
∼q∧r |
∼p∧(∼q∧r) |
(∼p∧(∼q∧r)⋁(q∧r) |
((∼p∧(∼q∧r))∨(q∧r)⋁(p∧r) |
((∼p∧(∼q∧r))∨(q∧r)⋁(p∧r)↔r |
T |
T |
T |
F |
F |
T |
T |
F |
T |
T |
T |
T |
T |
T |
F |
F |
F |
F |
F |
T |
F |
F |
F |
T |
T |
F |
T |
F |
T |
T |
F |
T |
F |
T |
T |
T |
T |
F |
F |
F |
T |
F |
T |
F |
T |
T |
T |
F |
F |
T |
T |
T |
F |
F |
T |
F |
F |
F |
F |
F |
F |
T |
F |
T |
F |
T |
F |
T |
T |
T |
T |
T |
F |
F |
T |
T |
T |
F |
F |
T |
T |
T |
T |
T |
F |
F |
F |
T |
T |
T |
T |
F |
F |
T |
T |
T |
So we Show that: ((∼p∧(∼q∧r))∨(q∧r)⋁(p∧r)↔r
the 2nd is wrong
ap ny Q2 kia hai
IDEA Solution file Attached
Q2 ANS IS WRONG
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