document.write( "Question 1112731: 1) (A&B) > ~(S v T)
\n" ); document.write( "2) ~E
\n" ); document.write( "3) (S v T) v ~(~E & ~F)
\n" ); document.write( "4) (~E v F) > (A&B)\r
\n" ); document.write( "\n" ); document.write( "Conclusion: E v F \n" ); document.write( "
Algebra.Com's Answer #728024 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Since ~E is given, the only way for the conclusion to be true is if F is true. Assume ~F. Assuming ~F and given ~E, (~E & ~F) must be true, hence ~(~E & ~F) is false. That leads to (S v T) true by Disjunctive Syllogism applied to (S v T) v ~(~E v ~F). But (S v T) true means ~(S v T) false, hence (A & B) is false by Modus Tollens applied to (A & B) -> ~(S v T). But (A & B) false means (~E v F) is false by Modus Tollens applied to (~E v F) -> (A & B). (~E v F) false means ~(~E v F) is true, which is the same as E & ~F by De Morgan. From E & ~F we get E by Conjunction Elimination (aka Simplification). But we were given ~E, so assuming ~F leads to the contradiction ~E & E. Hence ~F is false, hence F. Finally E v F by Disjunction Introduction (aka Addition)\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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