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How to prove ( q or not q), which is tautology when an arbitrary premise is given? (Using natural deduction)

I tried to make steps that finally goes to HS but failed.

Moibois
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Case 1: q is true, therefore q or x is true for any x, especially for x= not q Case 2: q is false, therefore not q is true, therefore x or not q is true for any x, especially for x=q Because by law of exclused middle, q has to be false or true, those cases are exhaustive.

Edit: q or not q= not(not q and not not q)= not (not q and q)= not (False)=True using excluded middle in step 2

axioman
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Not every proof system will be able to prove this without any assumption. And when you say 'natural deduction' ... that could mean quite a few different proof systems! Maybe you could let us know what rules of inference you have?

Bram28
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