document.write( "Question 980746: Rues of implication
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document.write( "I have been stuck on this for a while please help\r
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document.write( "~HvF
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document.write( "V&~K
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document.write( "I>A
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document.write( "~H>I
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document.write( "(VvN)>(F>~G)
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document.write( "/Av~G \n" );
document.write( "
Algebra.Com's Answer #601888 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "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.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "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.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The alternative is ~G, from which we can conclude A v ~G by addition.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Either way, A v ~G. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "My calculator said it, I believe it, that settles it\r \n" ); document.write( "\n" ); document.write( " |