MHB How Does De Morgan's Law Apply to Probability Logic?

Thread starter skaeno
Start date Sep 7, 2013
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Logic

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De Morgan's Law states that the disjunction of two propositions can be expressed as the negation of the conjunction of their negations. Given the equations τ(¬p)=1-τ(p) and τ(p∧q)=τ(p)∙τ(q), it can be shown that τ(p∨q) equals τ(p)+τ(q)-τ(p)∙τ(q). This relationship highlights the connection between probability logic and logical operations. The discussion emphasizes the importance of understanding these foundational principles in probability theory. Overall, De Morgan's Law provides a crucial framework for analyzing probabilities in logical expressions.

Sep 7, 2013

skaeno

If τ(¬p)=1-τ(p) and τ(p∧q)=τ(p)∙τ(q) show that:
τ(p∨q)=τ(p)+τ(q)-τ(p)∙τ(q)

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Sep 7, 2013

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skaeno said:

If τ(¬p)=1-τ(p) and τ(p∧q)=τ(p)∙τ(q) show that:
τ(p∨q)=τ(p)+τ(q)-τ(p)∙τ(q)

Hi skaeno, welcome to MHB! :)

Perhaps you might consider that p∨q=¬(¬p∧¬q)?

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