Can This Logic Proof for ~(~E > A) Be Resolved?

[NileQueen]

1. ~(A v B) It is neither A nor B.
2. ~(C v D) It is neither C nor D.
3. E > (C v B) If E, then C or B
Therefore, ~(~E > A)
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This is a logic proof I cannot seem to resolve.

I've tried working backwards from the conclusion.

http://www.hu.mtu.edu/~tlockha/h2701n10.s02.doc
Here are rules of replacement and rules of implication.

I know that ~ (~E > A) is a true statement.
I know that because I know ~E > A is false
Because ~E > ~ A (proved below) is true so ~E > A cannot be true.
A is not true because ~A is true. ~E is also true.

~(~E > A) cannot be unpacked because it is a horseshoe.
I can introduce (~E > A) as an addition (trying to work backwards):

1. ~A v (~E > A) add on ~A
2. A > (~E > A) 1, impl
We don't have a rule that says not P therefore not Q

3 (~E > A) v ~A 1 comm
4. ~ (~E >A) > ~A 3 impl

We don't have a rule that says Q therefore P


The proof as far as I can get it
> 1. ~(A v B)
> 2. ~(C v D)
> 3. E > (C v B)
> Therefore ~( ~E > A)
> 4. ~A and ~B ....1, DeMorgan
> 5. ~C and ~D ...2. DeMorgan
> 6. ~C...... 5. simplification
> 7. ~B......4. simplification
> 8. ~C and ~B...6,7 conjunction
> 9. ~(C v B)...8, DeMorgan
> 10. ~E.....3,9, Modus Tolens
11. ~A......4 simpl




Additional attempts:
> 12. ~A v ~A....11 tautology
> 13. A > ~A.....12. implication
> 14 ~A v E...11 addition
> 15. A> E.....14. impl
> 16. ~E > ~A....15. contraposition

17 ~A v ~E....11, add
18 A > ~E......17 impl
19 E > ~A......18, contra

 

[AKG]

In your first attempt, you get ~E (line 10) and ~A (line 11). This gives
12. ~E & ~A (conjunction introduction, 10, 11)
13. ~(~(~E & ~A)) (double negation, 12)
14. ~(~E > A) (implication, 13)

 

[NileQueen]

I agree with 12.

According to Demorgan
~E*~A :: ~(E v A)

Implication
(P>Q) :: (~P v Q)
no conjunctions are involved.

 

[NileQueen]

I think you have skipped at least one step.
Also, can you really operate on things within a parentheses with a negation outside?
~(E v A ) becomes ~ (~E >A) with DN and impl
I don't see that in the rules anywhere.
Thanks for your help.