Question 1191576
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I'm not sure how math_helper got that derivation. 
Unfortunately it's not correct because for instance Modus Tollens needs two components to it. 
Modus Tollens is of the form
P -> Q
~Q
Therefore ~P
However, math_helper only mentioned one line.
We would need the premise M to be able to get ~~(L & M) = L & M 


Also, there's a mistake in going from something like L -> (L & M) to L -> M using the simplification rule. 
It's not valid because the simplification rule must be applied to entire arguments only and not pieces of such. Unfortunately we can't go from L & M to L even though it's tempting to do so.


Here's how I would do the derivation
<table border = "1" cellpadding = "5"><tr><td>Number</td><td>Statement</td><td>Line(s) Used</td><td>Reason</td></tr><tr><td>1</td><td>( L & M ) v ~L</td><td></td><td></td></tr><tr><td>2</td><td>~(L & M) -> ~M</td><td></td><td></td></tr><tr><td>:.</td><td>L <--> M</td><td></td><td></td></tr><tr><td>3</td><td>(L v ~L) & (M v ~L)</td><td>1</td><td>Distribution</td></tr><tr><td>4</td><td>M v ~L</td><td>3</td><td>Simplification</td></tr><tr><td>5</td><td>~L v M</td><td>4</td><td>Commutation</td></tr><tr><td>6</td><td>L -> M</td><td>5</td><td>Material Implication</td></tr><tr><td>7</td><td>(L & M) v ~M</td><td>2</td><td>Material Implication</td></tr><tr><td>8</td><td>(L v ~M) & (M v ~M)</td><td>7</td><td>Distribution</td></tr><tr><td>9</td><td>L v ~M</td><td>8</td><td>Simplification</td></tr><tr><td>10</td><td>~M v L</td><td>9</td><td>Commutation</td></tr><tr><td>11</td><td>M -> L</td><td>10</td><td>Material Implication</td></tr><tr><td>12</td><td>(L -> M) & (M -> L)</td><td>6,11</td><td>Conjunction</td></tr><tr><td>13</td><td>L <--> M</td><td>12</td><td>Material Equivalence</td></tr></table>
Other derivations may be possible (and are probably likely).
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