Question 980746
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You need to do a conditional proof.  The first step is to note that you can deduce V from V & ~K by simplification.  Then by Addition (or Disjunction Introduction if you prefer), you can deduce V v N.  Then by Modus Ponens we conclude F -> ~G.


Now the conditional part.  Assume G.  Then by Modus Tollens you have ~F.  ~F and ~H v F gives us ~H.  Then by Modus Ponens, ~H -> I gives us I, then by Modus Ponens again, I -> A gives us A.  In summary, assuming G we get A and from A we can conclude A v ~G by Addition.


The alternative is ~G, from which we can conclude A v ~G by addition.


Either way, A v ~G. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

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