1. (L & M) v ~L          Premise

2. ~(L & M) --> ~M       Premise

// show  L <--> M



3.  L --> (L & M)        1, Material Implication (MImpl)

4.  M --> (L & M)        2, Modus Tollens (MT)

5.  L --> M              3, Simplification (SIMP)

6.  M --> L              4, SIMP

7.  L <--> M             5,6 Material Equivalence (MEquiv)



DONE

 

Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!
 

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


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


Other derivations may be possible (and are probably likely).

 

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