Question 1191576: please help me solve this: Construct a proof of the following argument: (L&M) ∨ ¬L, ¬(L&M) →
¬M ∴ L ↔ M.
Found 2 solutions by math_helper, math_tutor2020:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
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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