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The key to solving this kind of deduction is how to perform the disjunctive syllogism, i..e how get from A v B and ¬A to B, using disjunction elimination. The idea is the following: There two cases to consider -- either A or B. Assume A. This leads to a contradiction with the second assumption ¬A, so we get ⊥. From a contradiction we may assume anything, so ...


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Good day. I do not quite understand how I can get ~~p after the 11th line. Good day. You do not necessarily require ~~p to contradict ~p, you simply need some contradiction to be derivable from assuming ~(p | ~p). Well, (p | ~p) contradicts the assumption itself, so... Don't reiterate on line 10. The sub-proof is complete (for your purpose) at line 9. ...


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Here is a complete proof with all steps justified in the Fitch program.


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Your proof is valid but rather difficult to follow. It is quite disorganised. It would be clearer to collate the contexts together so that the reader does not have to track back and forth. (I also prefer using just one disjunction elimination, although using two is not wrong.) | _ 1 ( 1) A v (B & C) P || 2 ( 2) A H || 2 (...


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