document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=981965>Question 981965</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. Q > ~Q<BR>\n" );
document.write( "2. ~(G & Q) > U<BR>\n" );
document.write( "    / ~(G & ~U)\r<BR>\n" );
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document.write( "Can somebody solve this proof please? Thank you :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #602848 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=602848\">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 complete opposite of the conclusion. Assume (G & ~U)\r<BR>\n" );
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document.write( "Step 2) Use simplification to get ~U\r<BR>\n" );
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document.write( "Step 3) Use modus tollens with line 2 and ~U to get ~~(G & Q) which turns into (G & Q)\r<BR>\n" );
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document.write( "Step 4) Simplification frees up Q\r<BR>\n" );
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document.write( "Step 5) Modus ponens on line 1, and using the freed up Q from step 4, consequently frees up ~Q\r<BR>\n" );
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document.write( "Step 6) We have Q and ~Q they conjunct to (Q & ~Q) which is always false. This is a contradiction.\r<BR>\n" );
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document.write( "Since we have a contradiction, the initial assumption (G & ~U) is false which makes the opposite true. That proves ~(G & ~U) is a proper conclusion. </SPAN>\n" );
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