document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=981975>Question 981975</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) <BR>\n" );
document.write( "Can somebody solve this proof please? Thank you :)\r<BR>\n" );
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document.write( "\"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.\"\r<BR>\n" );
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document.write( "Jim, is there anyway to do this proof without using a contradiction and not assuming? 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 #602850 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=602850\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Alternative method\r<BR>\n" );
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document.write( "<tr><th>Number</th><th>Statement</th><th>Lines Used</th><th>Reason</th></tr><tr><TD >1</TD><TD >Q &gt; ~Q</TD><TD ></TD><TD ></TD></tr><tr><TD >2</TD><TD >~(G &amp; Q) &gt; U</TD><TD ></TD><TD ></TD></tr><tr><TD >:.</TD><TD >~(G &amp; ~U)</TD><TD ></TD><TD ></TD></tr><tr><TD >3</TD><TD >~Q v ~Q</TD><TD >1</TD><TD >Material Implication</TD></tr><tr><TD >4</TD><TD >~Q</TD><TD >3</TD><TD >Tautology</TD></tr><tr><TD >5</TD><TD >~~(G &amp; Q) v U</TD><TD >2</TD><TD >Material Implication</TD></tr><tr><TD >6</TD><TD >(G &amp; Q) v U</TD><TD >5</TD><TD >Double Negation</TD></tr><tr><TD >7</TD><TD >U v (G &amp; Q)</TD><TD >6</TD><TD >Commutation</TD></tr><tr><TD >8</TD><TD >(U v G) &amp; (U v Q)</TD><TD >7</TD><TD >Distribution</TD></tr><tr><TD >9</TD><TD >(U v Q) &amp; (U v G)</TD><TD >8</TD><TD >Commutation</TD></tr><tr><TD >10</TD><TD >U v Q</TD><TD >9</TD><TD >Simplification</TD></tr><tr><TD >11</TD><TD >U</TD><TD >10,4</TD><TD >Disjunctive Syllogism</TD></tr><tr><TD >12</TD><TD >U v ~G</TD><TD >11</TD><TD >Addition</TD></tr><tr><TD >13</TD><TD >~G v U</TD><TD >12</TD><TD >Commutation</TD></tr><tr><TD >14</TD><TD >~(~~G &amp; ~U)</TD><TD >13</TD><TD >De Morgan's Law</TD></tr><tr><TD >15</TD><TD >~(G &amp; ~U)</TD><TD >14</TD><TD >Double Negation</TD></tr>\n" );
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