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

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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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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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