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Prove ~(~A&~B) from A in as few lines as possible.

~ = negation
& = conjunction
v = disjunction
| = line in a subproof

Here's what I have:

  1. A - Premise
  2. |~A - Assume
  3. |~B - Assume
  4. |~A&~B - &Intro;3.4
  5. ~(~A&~B) - ~Intro;4

I'm quite sure this is wrong but I don't know how to fix it. Any help, even advice or tips, would be greatly appreciated!

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  • You have not discharged the two additional assumptions 2 and 3. Commented Feb 3, 2020 at 8:19
  • 1
    You have to start with premise A and with assumption (~A&~B). Commented Feb 3, 2020 at 8:20
  • I'm confused by how you can assume ~A when you have A as a premise. Also, don't you want to distribute ~ into the bracket? en.wikipedia.org/wiki/Negation#Distributivity en.wikipedia.org/wiki/De_Morgan%27s_laws
    – puppetsock
    Commented Feb 3, 2020 at 21:09
  • 1
    @puppetsock You may assume anything you wish for the sake of an argument. However, it is not useful to assume ~A. Commented Feb 4, 2020 at 3:05
  • If the answer below is enough for you, please accept it and we can "close" the post. Commented Feb 21, 2020 at 10:27

1 Answer 1

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No. Your subproof is drawkcab. You are not aiming to derive a position from a random assumption.

Negation introduction works by deriving a contradiction when assuming a position (~A & ~B), and thusly inferring its negation (~(~A & ~B)) holds when that assumption is discharged.

And so ...

  |_ A            premise
  |  |_ ~A & ~B   assumption
  |  |  :         :
  |  |  #         ~ elimination
  |  ~(~A & ~B)   ~ introduction

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