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 Post subject: Propositional Logic
PostPosted: Thu, 4 Feb 2010 14:50:33 UTC 
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I am having serious trouble with proving this sequent and would appreciate some help!

: ((P → ~ P) → P) → ((P → ~ P) → ~ P)

it must be achieved in ten steps, and is a theorem so has no premises

Thanks guys!!


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 Post subject: Re: Propositional Logic
PostPosted: Thu, 4 Feb 2010 19:04:45 UTC 
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MissSarahLee wrote:
I am having serious trouble with proving this sequent and would appreciate some help!

: ((P → ~ P) → P) → ((P → ~ P) → ~ P)

it must be achieved in ten steps, and is a theorem so has no premises

Thanks guys!!


Don't know exactly which rules you are allowed to use, but a way to prove the statement is first proving (p\Rightarrow\neg p)\Rightarrow(p\Rightarrow\neg p), then apply (p\Rightarrow(q\Rightarrow r))\Rightarrow((p\Rightarrow r)\Rightarrow(q\Rightarrow r)) and modus ponens.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Last edited by outermeasure on Thu, 4 Feb 2010 19:13:27 UTC, edited 1 time in total.

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 Post subject: hmm
PostPosted: Thu, 4 Feb 2010 19:10:16 UTC 
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Im not entirely sure about that, im not learning such advanced stuff yet

currently i am using the basic rules, such as & elimination and introduction, conditional proof, reductio ad absurdum, disjunction elimination and introduction, modus ponens, modus tolens, biconditional introduction and elimination and basic propositional logic like that, thanks for your reply though:)


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