User talk:Dante Cardoso Pinto de Almeida

Welcome
Good to see your contributions.

You seem to have hit the ground running, but I will give you the standard intro script, as follows:

Welcome
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Disjunction and Implication
An interesting theorem you've posted up! Natural deduction is fun. I'll take a look at it later - but I have a project on this weekend and the next few weeks, so I may not be able to get to it for some time.

Cheers, Matt Westwood (talk)

Something along the lines:

Proposition $(p \implies q) \implies q$. Assume $\neg (p \lor q)$ and so $\neg p \land \neg q$. So $\neg q$ and so $\neg (p \implies q)$ but that means $\neg q \implies p$ giving us our required contradiction $p \land \neg p$. --Matt Westwood 18:57, 10 April 2010 (UTC)