document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=978498>Question 978498</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help with proofs. Thank you!<BR>\n" );
document.write( "D &#8594;(B&#8594;C), B & &#8722;C &#9500; &#8722;D  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #599878 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=599878\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You can do a proof by contradiction.\r<BR>\n" );
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document.write( "Step 1) Assume the opposite of the conclusion. Assume ~~D which is equivalent to D\r<BR>\n" );
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document.write( "Step 2) Since we have D, we can use D -> (B -> C) and modus ponens to get B -> C\r<BR>\n" );
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document.write( "Step 3) Use simplification to go from B & ~C to just B. You can also free up ~C as well\r<BR>\n" );
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document.write( "Step 4) Use B and B -> C to get C (modus ponens again)\r<BR>\n" );
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document.write( "Step 5) The ~C and C contradict one another. So the original assumption is not possible. Therefore, the opposite of the assumption (D) is the only possibility.\r<BR>\n" );
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document.write( "So that's one way we can prove that ~D is the conclusion. </SPAN>\n" );
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