document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164909>Question 1164909</A>: <I><SPAN STYLE=\"word-break: break-all;\"> i have been having a hard time answering this rules of inference and rules of replacement, can you help me answer this?<BR>\n" );
document.write( "1)  ﬢ[P→ﬢ(Q۸R)]<BR>\n" );
document.write( "ﬢ[P→ﬢ<BR>\n" );
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document.write( "    ∴P۸(Q۸R)\r<BR>\n" );
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document.write( "2)    1. (P۷Q)→(R۸S)<BR>\n" );
document.write( "        2. P<BR>\n" );
document.write( "_________________<BR>\n" );
document.write( "        ∴ R\r<BR>\n" );
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document.write( "3)    1. (P۷Q) ۷ ﬢR<BR>\n" );
document.write( "        2. ﬢP ۸ ﬢQ<BR>\n" );
document.write( "__________________<BR>\n" );
document.write( "           ∴ﬢR ۷ S\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #853504 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2189)</B></A><img src=\"/images/stars/stars-13.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=853504\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Rules of inference and replacement can feel like a complex puzzle at first, but once you identify the \"shape\" of the statements, the rules become much clearer. Here are the step-by-step proofs for your three problems.\r<BR>\n" );
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document.write( "### 1) Prove: \r<BR>\n" );
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document.write( "**Given:** \r<BR>\n" );
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document.write( "| Step | Statement | Rule Used |<BR>\n" );
document.write( "| --- | --- | --- |<BR>\n" );
document.write( "| 1 |  | Premise |<BR>\n" );
document.write( "| 2 |  | **Material Implication (Impl)** on 1 |<BR>\n" );
document.write( "| 3 |  | **De Morgan's (DM)** on 2 |<BR>\n" );
document.write( "| 4 |  | **Double Negation (DN)** on 3 |\r<BR>\n" );
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document.write( "**Explanation:** To break into a negated conditional, we first turn the \"if-then\" into an \"or\" statement, then use De Morgan's to distribute the negation.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### 2) Prove: \r<BR>\n" );
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document.write( "**Given:** 1. <BR>\n" );
document.write( "2. \r<BR>\n" );
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document.write( "| Step | Statement | Rule Used |<BR>\n" );
document.write( "| --- | --- | --- |<BR>\n" );
document.write( "| 1 |  | Premise |<BR>\n" );
document.write( "| 2 |  | Premise |<BR>\n" );
document.write( "| 3 |  | **Addition (Add)** on 2 |<BR>\n" );
document.write( "| 4 |  | **Modus Ponens (MP)** on 1, 3 |<BR>\n" );
document.write( "| 5 |  | **Simplification (Simp)** on 4 |\r<BR>\n" );
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document.write( "**Explanation:** Since we have , we can \"add\" anything to it using an \"or\" statement. This satisfies the left side of the conditional in Step 1, allowing us to extract the right side.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### 3) Prove: \r<BR>\n" );
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document.write( "**Given:**\r<BR>\n" );
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document.write( "1. <BR>\n" );
document.write( "2. \r<BR>\n" );
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document.write( "| Step | Statement | Rule Used |<BR>\n" );
document.write( "| --- | --- | --- |<BR>\n" );
document.write( "| 1 |  | Premise |<BR>\n" );
document.write( "| 2 |  | Premise |<BR>\n" );
document.write( "| 3 |  | **De Morgan's (DM)** on 2 |<BR>\n" );
document.write( "| 4 |  | **Disjunctive Syllogism (DS)** on 1, 3 |<BR>\n" );
document.write( "| 5 |  | **Addition (Add)** on 4 |\r<BR>\n" );
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document.write( "**Explanation:** In Step 3, we used De Morgan's to show that neither  nor  is true. This contradicts the first part of Premise 1, which forces  to be true. Once we have , we can use Addition to add any variable (in this case, ).\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Logic Tool Reference\r<BR>\n" );
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document.write( "A helpful tip for these is to always look at your **Conclusion** first. If it has a \"\" (like in problem 3), you often only need to find one side of it and use **Addition**.\r<BR>\n" );
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document.write( "Would you like to try another set, or should we go deeper into how **De Morgan's Law** works? </SPAN>\n" );
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