Question 1090173
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To do a reductio ad absurdum argument, I'm going to use a proof by contradiction.
Proof By Contradiction:


This is an informal paragraph style proof:


Assume the opposite of the conclusion is true
The conclusion is ~A v B
The opposite of the conclusion is ~(~A v B) = ~~A & ~B = A & ~B through the use of De Morgan's Law
If we assume A & ~B is true, then A is certainly true and so is ~B (<font color=blue>Keep the fact that ~B is true in mind</font>). Both parts of a conjunction must be true if the whole thing is true.
If A is the case, then so is ~~A
By modus tollens, we can arrive that ~Q is also the case.
Then through modus ponens, we can use ~Q and ~Q -> (L -> F) to find that L -> F is the case
Now use the premise L and L -> F to find that F is true (use modus ponens again)
Finally use F and F -> B to find B is true (another application of modus ponens)
But wait, earlier I said that ~B was true (<font color=blue>In a previous note above</font>). 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.
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. 


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Here's a more formal way to do the proof using a derivation table
<table border=1 cellpadding=3><tr><th colspan="2">Number</th><th>Statement</th><th>Lines Used</th><th>Reason</th></tr><tr><td>1</td><td></td><td>~Q -&gt; (L -&gt; F)</td><td></td><td></td></tr><tr><td>2</td><td></td><td>Q -&gt; ~A</td><td></td><td></td></tr><tr><td>3</td><td></td><td>F -&gt; B</td><td></td><td></td></tr><tr><td>4</td><td></td><td>L</td><td></td><td></td></tr><tr><td colspan="2">:.</td><td>~A v B</td><td></td><td></td></tr><tr><td></td><td>5</td><td>~(~A v B)</td><td></td><td>Assumption for Indirect Proof</td></tr><tr><td></td><td>6</td><td>~~A &amp; ~B</td><td>5</td><td>De Morgan's Law</td></tr><tr><td></td><td>7</td><td>A &amp; ~B</td><td>6</td><td>Double Negation</td></tr><tr><td></td><td>8</td><td>A</td><td>7</td><td>Simplication</td></tr><tr><td></td><td>9</td><td>~B</td><td>7</td><td>Simplication</td></tr><tr><td></td><td>10</td><td>~~A</td><td>8</td><td>Double Negation</td></tr><tr><td></td><td>11</td><td>~Q</td><td>2,10</td><td>Modus Tollens</td></tr><tr><td></td><td>12</td><td>L -&gt; F</td><td>1,11</td><td>Modus Ponens</td></tr><tr><td></td><td>13</td><td>F</td><td>12,4</td><td>Modus Ponens</td></tr><tr><td></td><td>14</td><td>B</td><td>3,13</td><td>Modus Ponens</td></tr><tr><td></td><td>15</td><td>B &amp; ~B</td><td>14,9</td><td>Conjunction</td></tr><tr><td>16</td><td></td><td>~A v B</td><td>5-15</td><td>Indirect Proof</td></tr></table>

Note: "Indirect Proof" is another term for "Proof by Contradiction"

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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. 


An informal proof would go like this:


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. 
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.
We use ~Q to get L -> F (modus ponens)
We use L and L -> F to get F (modus ponens)
We use F and F -> B to get B (modus ponens)
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.
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


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Here's a formal derivation table
<table border=1 cellpadding=3><tr><th colspan="2">Number</th><th>Statement</th><th>Lines Used</th><th>Reason</th></tr><tr><td>1</td><td></td><td>~Q -&gt; (L -&gt; F)</td><td></td><td></td></tr><tr><td>2</td><td></td><td>Q -&gt; ~A</td><td></td><td></td></tr><tr><td>3</td><td></td><td>F -&gt; B</td><td></td><td></td></tr><tr><td>4</td><td></td><td>L</td><td></td><td></td></tr><tr><td colspan="2">:.</td><td>~A v B</td><td></td><td></td></tr><tr><td></td><td>5</td><td>A</td><td></td><td>Assumption for Conditional Proof</td></tr><tr><td></td><td>6</td><td>~~A</td><td>5</td><td>Double Negation</td></tr><tr><td></td><td>7</td><td>~Q</td><td>2,6</td><td>Modus Tollens</td></tr><tr><td></td><td>8</td><td>L -&gt; F</td><td>1,7</td><td>Modus Ponens</td></tr><tr><td></td><td>9</td><td>F</td><td>8,4</td><td>Modus Ponens</td></tr><tr><td></td><td>10</td><td>B</td><td>3,9</td><td>Modus Ponens</td></tr><tr><td>11</td><td></td><td>A -&gt; B</td><td>5-10</td><td>Conditional Proof</td></tr><tr><td>12</td><td></td><td>~A v B</td><td>11</td><td>Material Implication</td></tr></table>
Note how this derivation table has a lot in common with the previous table.
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