- #1

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## 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

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- Thread starter aeronautical
- Start date

In summary, the conversation discusses determining if the expression ((p → q) Ʌ (q → p) → (p Ʌ q) is a tautology and creating a table to evaluate its truth value for all possible values of p and q. The table is used to show that the expression is true for all combinations of p and q, therefore making it a tautology.

- #1

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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

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- #2

Homework Helper

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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

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Cyosis said:

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 I am lost. this is very tricky. i don't understand this table

- #4

Homework Helper

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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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