document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1135774>Question 1135774</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help! I can't for the life of me figure this one out!\r<BR>\n" );
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document.write( "Proof by natural deduction- Predicate Logic. Use a direct proof to show that the following argument is valid.\r<BR>\n" );
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document.write( "Premise 1: (∃x)Kx --> (x) (Lx --> Mx)<BR>\n" );
document.write( " Premise 2: Kc • Lc<BR>\n" );
document.write( " Conclusion: Mc\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #753448 by </B><A HREF=/tutors/aboutme.mpl?userid=MathLover1><B>MathLover1(20850)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=MathLover1><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=MathLover1><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=753448\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hint : assuming you are using Natural Deduction\r<BR>\n" );
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document.write( "From Premise-2 : Kc∧Lc\r<BR>\n" );
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document.write( "you have to derive Kc (by ∧-elimination), followed by ∃xKx (by ∃-introduction).\r<BR>\n" );
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document.write( "In this way, you can use → -elimination (i.e. modus ponens) with Premise-1 and derive : (∀x)(Lx→Mx).\r<BR>\n" );
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document.write( "Now you have to use ∀-elimination (i.e. universal instantiation) with c to get : Lc→Mc. </SPAN>\n" );
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