document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=251007>Question 251007</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There is only one premise, and the conclusion has different letters\r<BR>\n" );
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document.write( "Here it is:\r<BR>\n" );
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document.write( "A -> B\r<BR>\n" );
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document.write( "We have to show (A & C) -> B </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #185034 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=185034\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If we use the addition property, we go from A -> B to (A -> B) V ~C (the \"V\" stands for 'or' and \"~\" stands for 'not')\r<BR>\n" );
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document.write( "Rearrange the terms to get ~C V (A -> B) and then use material implication to get C -> (A -> B). From here, use exportation to get (C & A) -> B and then use commutation to get (A & C) -> B. This derivation is pretty straightforward, but the only trick here is the addition of ~C. </SPAN>\n" );
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