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Show that each of the conditional statement is a tautology
not using truth table

 

[(p → q) ∧ (q → r)] → (p → r)
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By Tautology we mean that every output should be True.

We will be proving this by contradiction.

For implication to be False there is only one condition which is T->F

1-let  p->r == False This implies p==T and r==F

2-According to our assumption  [(p → q) ∧ (q → r)]  should be True and we have  p==T and r==F

3-For  [(p → q) ∧ (q → r)]  to be True  both (p → q) and  (q → r) should be true according to the property of AND

4- For (p → q) to be true q should be True 

5-Since q==True  (q → r) becomes False  (T->F)

6-because of 5   [(p → q) ∧ (q → r)] becomes False 

This contradicts our assumption of  [(p → q) ∧ (q → r)] → (p → r) Being False 

So we can conclude that this can never be False

Hence It is a Tautology .

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[(p → q) ∧ (q → r)] → (p → r)

(p → q)=P'+Q== region(2,3,6,8)+region (2,4,6,7)=region (2,3,4,6,7,8) .. (Here + is equivalent to logical OR)

Q->R =Q'+R== region(1,3,5,6,7,8)

P->R=P'+R== region(2,3,5,6,7,8)

[(p → q) ∧ (q → r)]  == Region (3,6,7,8) .....1

(p → r) ........2

[(p → q) ∧ (q → r)] → (p → r) == 1->2== 1'+2 == Region(1,2,4,5)+Region(2,3,5,6,7,8)=Region (1,2,3,4,5,6,7,8) Hence Tautology

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(p' +q)(q'+r)->(p'+r)

[p'q'+p'r+qr]->(p'+r)

pq'+pr'q'+pr'+pr'+qpr'+qr'+p'+r

pq' +pr'+qr'+p'+r

p' + q' +pr' + qr' + r

p' + q' + pr' + q +r

p' + 1 + pr' + r

1+r+p

= 1 (true)

solved using properties of boolean algebra

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