document.write( "Question 1028017: I'm having a hard time constructing a proof for the following argument.\r
\n" ); document.write( "\n" ); document.write( "M&~Q
\n" ); document.write( "M->S
\n" ); document.write( "S->P
\n" ); document.write( "And the goal is ~(P->Q)
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Algebra.Com's Answer #643294 by richard1234(7193)\"\" \"About 
You can put this solution on YOUR website!
M is true, so apply rules 2 and 3:\r
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\n" ); document.write( "\n" ); document.write( "M --> S --> P, so P is true
\n" ); document.write( "Q is not true.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Therefore it is not true that P implies Q.
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