document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1112710>Question 1112710</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Indirect proof<BR>\n" );
document.write( "9. <BR>\n" );
document.write( "1) R<BR>\n" );
document.write( "2) (~ C v ~ D) v S<BR>\n" );
document.write( "3) ~ (C &#8901; D) &#8835; ~R		/ &#8756; S\r<BR>\n" );
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document.write( "10. <BR>\n" );
document.write( "1) (A &#8901; B)&#8901; ~ (S v T)<BR>\n" );
document.write( "2) ~E<BR>\n" );
document.write( "3) (S v T) v ~ (~E &#8901; ~F)<BR>\n" );
document.write( "4) (~E v F)&#8835;(A &#8901; B)		/ &#8756; E v F\r<BR>\n" );
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document.write( "12.<BR>\n" );
document.write( "1) A v B<BR>\n" );
document.write( "2) B &#8835; (A v D)<BR>\n" );
document.write( "3) ~ D				/ &#8756; A\r<BR>\n" );
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document.write( "Conditional proof<BR>\n" );
document.write( "11.<BR>\n" );
document.write( "1) A &#8835; B			/ &#8756; A &#8835; [ C &#8835; ~ (B &#8835; ~A) ]\r<BR>\n" );
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document.write( "Help me please! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #727861 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=727861\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "9. Indirect Proof\r<BR>\n" );
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document.write( "You can format this however you want but the basic idea is to assume the negation of the desired conclusion and have that lead you to a contradiction, that is a statement that is identically false.\r<BR>\n" );
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document.write( "Assume ~S.  Then from (2), you get ~C v ~D by Disjunctive Syllogism.  Then you can write ~(C & D) from the previous statement using De Morgan.  From this and (3) you get ~R by Modus Ponens.  But that leads to R & ~R which is identically false.  Hence the assumption, ~S, is false, therefore S.\r<BR>\n" );
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document.write( "One question per post in accordance with the posting instructions, please.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \\r\"> </SPAN>\n" );
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