document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1178946>Question 1178946</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. ~(K v F)<BR>\n" );
document.write( "2. ~F ⊃ (K v C)<BR>\n" );
document.write( "3. (G v C) ⊃ ~H		/~(K v H)<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 #850278 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850278\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's analyze this logical argument to determine its validity. We'll use a proof by deduction to see if we can derive ~(K v H) from the given premises.\r<BR>\n" );
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document.write( "**1. Premises:**\r<BR>\n" );
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document.write( "1.  ~(K v F)<BR>\n" );
document.write( "2.  ~F ⊃ (K v C)<BR>\n" );
document.write( "3.  (G v C) ⊃ ~H\r<BR>\n" );
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document.write( "**2. Derivations:**\r<BR>\n" );
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document.write( "4.  ~K • ~F (De Morgan's Law on premise 1)<BR>\n" );
document.write( "5.  ~F (Simplification on line 4)<BR>\n" );
document.write( "6.  K v C (Modus Ponens on lines 2 and 5)<BR>\n" );
document.write( "7.  ~K (Simplification on line 4)<BR>\n" );
document.write( "8.  C (Disjunctive Syllogism on lines 6 and 7)<BR>\n" );
document.write( "9.  G v C (Addition on line 8)<BR>\n" );
document.write( "10. ~H (Modus Ponens on lines 3 and 9)<BR>\n" );
document.write( "11. ~K • ~H (Conjunction on lines 7 and 10)<BR>\n" );
document.write( "12. ~(K v H) (De Morgan's Law on line 11)\r<BR>\n" );
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document.write( "**Explanation of Steps:**\r<BR>\n" );
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document.write( "* **Step 4:** We applied De Morgan's Law to premise 1. De Morgan's Law states that ~(A v B) is equivalent to ~A • ~B.<BR>\n" );
document.write( "* **Step 5:** We used simplification on line 4. If a conjunction (A • B) is true, then both A and B are true individually.<BR>\n" );
document.write( "* **Step 6:** We used Modus Ponens on lines 2 and 5. Modus Ponens states that if (A ⊃ B) is true and A is true, then B is true.<BR>\n" );
document.write( "* **Step 7:** We used simplification on line 4, just as in step 5.<BR>\n" );
document.write( "* **Step 8:** We used Disjunctive Syllogism on lines 6 and 7. Disjunctive Syllogism states that if (A v B) is true and ~A is true, then B is true.<BR>\n" );
document.write( "* **Step 9:** We used addition on line 8. If A is true, then (A v B) is also true, regardless of the truth value of B.<BR>\n" );
document.write( "* **Step 10:** We used Modus Ponens on lines 3 and 9. Modus Ponens states that if (A ⊃ B) is true and A is true, then B is true.<BR>\n" );
document.write( "* **Step 11:** We used conjunction on lines 7 and 10. If A and B are both true, then (A • B) is true.<BR>\n" );
document.write( "* **Step 12:** We applied De Morgan's Law to line 11. De Morgan's Law states that ~A • ~B is equivalent to ~(A v B).\r<BR>\n" );
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document.write( "**Conclusion:**\r<BR>\n" );
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document.write( "We were able to derive ~(K v H) from the given premises. Therefore, the argument is **valid**.<BR>\n" );
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