document.write( "Question 374237: I am having trouble solving the following propositional logic proofs. \r
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document.write( "1. Conclusion: N + Y
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document.write( "1. K > (N & P)
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document.write( "2. S > (X & Y)
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document.write( "3. S + K
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document.write( "2. Conclusion: ~Q > ~M
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document.write( "1. M > [ Q + ~(E + H) ]
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document.write( "2. ~E > H
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document.write( "3. Conclusion: A & D
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document.write( "1. ~D > (F & P)
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document.write( "2. ~A > ~(F + P)
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document.write( "3. (F <=> P) > (~~D & A)
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document.write( "4. Conclusion: F > (Y > F)
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document.write( "1. Q
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document.write( "5. Conclusion: ~(~Q > ~X) > ~J
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document.write( "1.J > (M > Q)
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document.write( "2. X > (M + V)
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document.write( "3. J > (V > ~X) \n" );
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Algebra.Com's Answer #266334 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! I'll do the first one to get you started.\r \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( "1. K > (N & P)\r\n" ); document.write( "2. S > (X & Y)\r\n" ); document.write( "3. S + K :. N+Y\r\n" ); document.write( "-------------------------------------------------\r\n" ); document.write( "4. [K > (N & P)] & [S > (X & Y)] 1,2 Conjunction\r\n" ); document.write( "5. (N & P) + (X & Y) 4,3 Constructive Dilemma\r\n" ); document.write( "6. [(N & P) + X] & [(N & P) + Y] 5 Distibution\r\n" ); document.write( "7. [(N & P) + Y] & [(N & P) + X] 6 Commutation\r\n" ); document.write( "8. (N & P) + Y 7 Simplification\r\n" ); document.write( "9. Y + (N & P) 8 Commutation\r\n" ); document.write( "10. (Y + N) & (Y + P) 9 Distribution\r\n" ); document.write( "11. Y + N 10 Simplification\r\n" ); document.write( "12. N + Y 11 Commutation\r\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |