document.write( "Question 1046573: please help me by showing the proof of the following using the rules of replacements..This propositions are logically equivalences and I need to see the proofs..thank you..
\n" ); document.write( "1. (P v Q)=>R = (P=>R)^(Q=>R)
\n" ); document.write( "2. P v (P^Q) = P
\n" ); document.write( "3. P ^ (PvQ) = P \n" ); document.write( "
Algebra.Com's Answer #662025 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
1. (P v Q)=>R => (P=>R)^(Q=>R)
\n" ); document.write( "Proof:
\n" ); document.write( " (P v Q)=>R
\n" ); document.write( "---> ~(P v Q) V R (Material implication)
\n" ); document.write( "---> (~P ^ ~Q) v R (de Morgan's)
\n" ); document.write( "---> (~P v R) ^ (~Q v R) (distributivity)
\n" ); document.write( "---> (P=>Q) ^ (Q=>R) (Material implication)\r
\n" ); document.write( "\n" ); document.write( "To prove (P=>R)^(Q=>R) => (P v Q)=>R, just reverse the steps in the preceding argument.\r
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\n" ); document.write( "\n" ); document.write( "2. By addition, it is easy to see that P => P v (P^Q) .
\n" ); document.write( "To prove P v (P^Q) => P:\r
\n" ); document.write( "\n" ); document.write( "P v (P^Q)
\n" ); document.write( "---> (PvP)^(PvQ) (distributivity)\r
\n" ); document.write( "\n" ); document.write( "---> P^(PvQ) (idempotency)
\n" ); document.write( "---> P (simplification)\r
\n" ); document.write( "\n" ); document.write( "Therefore, P v (P^Q) = P.\r
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\n" ); document.write( "\n" ); document.write( "3. I leave this up to you.\r
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