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\n" ); document.write( "To do a reductio ad absurdum argument, I'm going to use a proof by contradiction.
\n" ); document.write( "Proof By Contradiction:\r
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\n" ); document.write( "\n" ); document.write( "This is an informal paragraph style proof:\r
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\n" ); document.write( "\n" ); document.write( "Assume the opposite of the conclusion is true
\n" ); document.write( "The conclusion is ~A v B
\n" ); document.write( "The opposite of the conclusion is ~(~A v B) = ~~A & ~B = A & ~B through the use of De Morgan's Law
\n" ); document.write( "If we assume A & ~B is true, then A is certainly true and so is ~B (Keep the fact that ~B is true in mind). Both parts of a conjunction must be true if the whole thing is true.
\n" ); document.write( "If A is the case, then so is ~~A
\n" ); document.write( "By modus tollens, we can arrive that ~Q is also the case.
\n" ); document.write( "Then through modus ponens, we can use ~Q and ~Q -> (L -> F) to find that L -> F is the case
\n" ); document.write( "Now use the premise L and L -> F to find that F is true (use modus ponens again)
\n" ); document.write( "Finally use F and F -> B to find B is true (another application of modus ponens)
\n" ); document.write( "But wait, earlier I said that ~B was true (In a previous note above). So how can B also be true at the same time? This is where the contradiction lies. Therefore, the expression ~(~A v B) cannot be true so the original ~A v B must be true.
\n" ); document.write( "So in short, we've assumed a condition -- assumed that ~(~A v B) was true -- but it led to an absurdity of B and ~B being true at the same time. \r
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\n" ); document.write( "\n" ); document.write( "Here's a more formal way to do the proof using a derivation table
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| Number | Statement | Lines Used | Reason | |
|---|---|---|---|---|
| 1 | ~Q -> (L -> F) | |||
| 2 | Q -> ~A | |||
| 3 | F -> B | |||
| 4 | L | |||
| :. | ~A v B | |||
| 5 | ~(~A v B) | Assumption for Indirect Proof | ||
| 6 | ~~A & ~B | 5 | De Morgan's Law | |
| 7 | A & ~B | 6 | Double Negation | |
| 8 | A | 7 | Simplication | |
| 9 | ~B | 7 | Simplication | |
| 10 | ~~A | 8 | Double Negation | |
| 11 | ~Q | 2,10 | Modus Tollens | |
| 12 | L -> F | 1,11 | Modus Ponens | |
| 13 | F | 12,4 | Modus Ponens | |
| 14 | B | 3,13 | Modus Ponens | |
| 15 | B & ~B | 14,9 | Conjunction | |
| 16 | ~A v B | 5-15 | Indirect Proof |
\n" ); document.write( "\n" ); document.write( "Note: \"Indirect Proof\" is another term for \"Proof by Contradiction\"\r
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\n" ); document.write( "If you want to use a conditional proof, then you first need to realize that ~A v B is logically equivalent to A -> B through the material implication rule. \r
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\n" ); document.write( "\n" ); document.write( "An informal proof would go like this:\r
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\n" ); document.write( "\n" ); document.write( "Start by assuming A. We can't directly jump to B with A since that would be too easy and we cannot use the conclusion as part of the premises. That would lead to cicular reasoning.
\n" ); document.write( "Instead turn A into ~~A (double negation). That would allow us to pull out ~Q by modus tollens. As you can probably guess by this point, the steps are very similar to those shown above.
\n" ); document.write( "We use ~Q to get L -> F (modus ponens)
\n" ); document.write( "We use L and L -> F to get F (modus ponens)
\n" ); document.write( "We use F and F -> B to get B (modus ponens)
\n" ); document.write( "This is where the proof differs than the section above. Instead of a contradiction, we have essentially arrived at the proper conclusion we want based on the assumption provided.
\n" ); document.write( "Basically we started with A and we did a bunch of logical steps to arrive at B. If we assume A is true, then somewhere down the line B is true. So naturally if A, then B follows. That is written as A -> B which is equivalent to ~A v B\r
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\n" ); document.write( "\n" ); document.write( "Here's a formal derivation table
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| Number | Statement | Lines Used | Reason | |
|---|---|---|---|---|
| 1 | ~Q -> (L -> F) | |||
| 2 | Q -> ~A | |||
| 3 | F -> B | |||
| 4 | L | |||
| :. | ~A v B | |||
| 5 | A | Assumption for Conditional Proof | ||
| 6 | ~~A | 5 | Double Negation | |
| 7 | ~Q | 2,6 | Modus Tollens | |
| 8 | L -> F | 1,7 | Modus Ponens | |
| 9 | F | 8,4 | Modus Ponens | |
| 10 | B | 3,9 | Modus Ponens | |
| 11 | A -> B | 5-10 | Conditional Proof | |
| 12 | ~A v B | 11 | Material Implication |
\n" ); document.write( "Note how this derivation table has a lot in common with the previous table.
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