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Show if this is a Tautology

  • #1

Homework Statement



Determine if the following is a tautology:

((p → q) Ʌ (q → p) → (p Ʌ q)

I don´t know how to show this. Can somebody pls show me all the steps
 

Answers and Replies

  • #2
Cyosis
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Make the following table:

p|q|p → q|q → p|(p → q) Ʌ (q → p)|p Ʌ q|((p → q) Ʌ (q → p) → (p Ʌ q)
T|T|
T|F|
F|T|
F|F|

Now finish this table, if the last column yields true for all possible values for p and q then ((p → q) Ʌ (q → p) → (p Ʌ q) is a tautology.
 
  • #3
Make the following table:

p|q|p → q|q → p|(p → q) Ʌ (q → p)|p Ʌ q|((p → q) Ʌ (q → p) → (p Ʌ q)
T|T|
T|F|
F|T|
F|F|

Now finish this table, if the last column yields true for all possible values for p and q then ((p → q) Ʌ (q → p) → (p Ʌ q) is a tautology.
im sorry but im lost. this is very tricky. i dont understand this table
 
  • #4
Cyosis
Homework Helper
1,495
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I have basically chopped your original expression into chunks. Every column of the table has a term of the original expression in it and I have used | to separate the columns. The last column has the entire expression in it.

Lets finish the first row:
p and q are true so q->p is true and p->q thus (p → q) Ʌ (q → p) is true. On the other hand we have p Ʌ q which is true so now we have all components that we want. So we can conclude, since (p → q) Ʌ (q → p) is true and (p Ʌ q) is true, ((p → q) Ʌ (q → p) → (p Ʌ q) must be true.

Now try to work your way through the other values of initial p and q. Note that I put all possible combinations in the first two columns.
 

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