document.write( "Question 1206425: Premise:
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\n" ); document.write( "\n" ); document.write( "Use either indirect proof or conditional proof (or both) and the eighteen rules of inference to derive the conclusion of the following symbolized argument. \n" ); document.write( "
Algebra.Com's Answer #843960 by mccravyedwin(408)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "This student has informed me that she is not allowed to use material\r\n" );
document.write( "implication as I did in the above.\r\n" );
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document.write( "(P ⊃ Q) <=> (~P ∨ Q)\r\n" );
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document.write( "Why teachers don't just prove this early on is beyond me.  It makes\r\n" );
document.write( "proofs a lot simpler.  \r\n" );
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document.write( "They wouldn't even need to prove equivalence, they could just prove\r\n" );
document.write( "(P ⊃ Q) ⊃ (~P ∨ Q), and that would make lots of proofs easier.  It's \r\n" );
document.write( "easy to prove indirectly, anyway.\r\n" );
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document.write( "First we prove \r\n" );
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document.write( "premise \r\n" );
document.write( "1. P ⊃ Q      conclusion ~P ∨ Q \r\n" );
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document.write( "                 |2. ~(~P ∨ Q)      Assumption for Indirect Proof\r\n" );
document.write( "                 |3. ~~P & ~Q       2, DeMorgan's law\r\n" );
document.write( "                 |4. P & ~Q         3, Double negation\r\n" );
document.write( "                 |5. P              4, Simplification\r\n" );
document.write( "                 |6. Q            1,5, Modus Ponens\r\n" );
document.write( "                 |7. ~Q & P         4, commutation\r\n" );
document.write( "                 |8. ~Q             7, Simplification\r\n" );
document.write( "                 |9. Q & ~Q       6,8, Conjunction\r\n" );
document.write( "~P ∨ Q     lines 2-9  Indirect proof.\r\n" );
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document.write( "Therefore (P ⊃ Q) ⊃ (~P ∨ Q)\r\n" );
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document.write( "Now we reverse the conclusion and premise:\r\n" );
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document.write( "premise \r\n" );
document.write( "1. ~P ∨ Q      conclusion P ⊃ Q \r\n" );
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document.write( "                 |2. P      Assumption for Conditional Proof\r\n" );
document.write( "                 |3. ~~P            2, Double negation\r\n" );
document.write( "                 |4. Q            1,3, Disjunctive syllogism\r\n" );
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document.write( "5. P ⊃ Q    lines 2-4      Conditional proof.\r\n" );
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document.write( "Therefore (~P ∨ Q) ⊃ (P ⊃ Q)\r\n" );
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document.write( "So we have proved\r\n" );
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document.write( "[(P ⊃ Q) ⊃ (~P ∨ Q)] & [(~P ∨ Q) ⊃ (P ⊃ Q)]\r\n" );
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document.write( "thus they are equivalent.  But all we need is the first part,\r\n" );
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document.write( "(P ⊃ Q) ⊃ (~P ∨ Q)\r\n" );
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document.write( "Edwin

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