1. ~(A v B) It is neither A nor B.(adsbygoogle = window.adsbygoogle || []).push({});

2. ~(C v D) It is neither C nor D.

3. E > (C v B) If E, then C or B

Therefore, ~(~E > A)

--------------------------------------------

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 [Broken]

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

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# Homework Help: Logic Proof

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