document.write( "Question 1191576: please help me solve this: Construct a proof of the following argument: (L&M) ∨ ¬L, ¬(L&M) →
\n" ); document.write( "¬M ∴ L ↔ M. \n" ); document.write( "

Algebra.Com's Answer #823889 by math_tutor2020(3816) $\"\"$ $\"About$

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\n" ); document.write( "I'm not sure how math_helper got that derivation.
\n" ); document.write( "Unfortunately it's not correct because for instance Modus Tollens needs two components to it.
\n" ); document.write( "Modus Tollens is of the form
\n" ); document.write( "P -> Q
\n" ); document.write( "~Q
\n" ); document.write( "Therefore ~P
\n" ); document.write( "However, math_helper only mentioned one line.
\n" ); document.write( "We would need the premise M to be able to get ~~(L & M) = L & M \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Also, there's a mistake in going from something like L -> (L & M) to L -> M using the simplification rule.
\n" ); document.write( "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.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Here's how I would do the derivation
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "

Number	Statement	Line(s) Used	Reason
1	( L & M ) v ~L
2	~(L & M) -> ~M
:.	L <--> M
3	(L v ~L) & (M v ~L)	1	Distribution
4	M v ~L	3	Simplification
5	~L v M	4	Commutation
6	L -> M	5	Material Implication
7	(L & M) v ~M	2	Material Implication
8	(L v ~M) & (M v ~M)	7	Distribution
9	L v ~M	8	Simplification
10	~M v L	9	Commutation
11	M -> L	10	Material Implication
12	(L -> M) & (M -> L)	6,11	Conjunction
13	L <--> M	12	Material Equivalence

\n" ); document.write( "Other derivations may be possible (and are probably likely).
\n" ); document.write( " \n" ); document.write( "

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