Logic (Sentence Negation) Please note that while this is a homework question, I'm simply asking for fact checking and explanations to my solution.
Sentence: If the file is not damaged and the processor is fast, then the printer is slow
My Solution:
Symbolic Translation:  $(F'∧ P) → S$
Negation: 


*

*$((F'∧ P) → S)'$

*$(F ∧ P') → S'$

*$((F ∧ P') → (S'))'$

*$(F' ∨ P) ∧ S$         --could the negation end here?

*$((F')' ∨ P') ∧ S$

*$(F ∨ P') ∧ S$  


I'm kind of stuck from fixing the errors highlighted by Hunan and Ross. I'm not sure if I should proceed with steps 5 and 6
 A: I check the following statements statement


*

*$((F'∧ P) → S)'$

*$(F ∧ P') → S'$ 


by the truth table generator by Samuel Williams and by John Halleck's Expression Evaluator.
I translate them in a slighly different notation that can be understood by the truth table generator


*

*~((~F& P) -> S)

*(F & ~P) -> ~S


First put  ~((~F& P) -> S)  in the generator. You get
            !
F  P  S  |  ~  (  (  ~  F  &  P  )  ->  S  )
--------------------------------------------
0  0  0  |  0        1  0  0  0      1  0     
0  0  1  |  0        1  0  0  0      1  1     
0  1  0  |  1        1  0  1  1      0  0     
0  1  1  |  0        1  0  1  1      1  1     
1  0  0  |  0        0  1  0  0      1  0     
1  0  1  |  0        0  1  0  0      1  1   
1  1  0  |  0        0  1  0  1      1  0     
1  1  1  |  0        0  1  0  1      1  1

The result for the expression is ion the column I marked with !
Now do the same with the second expression  (F & ~P) > ~S 
                               !
F  P  S  |  (  F  &  ~  P  )  ->  ~  S
--------------------------------------
0  0  0  |     0  0  1  0      1  1  0  
0  0  1  |     0  0  1  0      1  0  1  
0  1  0  |     0  0  0  1      1  1  0  
0  1  1  |     0  0  0  1      1  0  1  
1  0  0  |     1  1  1  0      1  1  0  
1  0  1  |     1  1  1  0      0  0  1  
1  1  0  |     1  0  0  1      1  1  0  
1  1  1  |     1  0  0  1      1  0  1  

Th column marked with ! differ so you made an error. Only if both columns match your derivation is right.
For the evaluator we need the following translation


*

*~((~F& P) > S)

*(F & ~P) > ~S


We put   (~((~F& P) > S))=((F & ~P) > ~S)  in the evaluator and get the result

We find that it is contingent, for example:
  
  
*
  
*(p=T s=F ) gives a true evaluation.  
  
*(p=F s=F ) gives a false evaluation.  
  

If the expression is TRUE the your derivation is ok, if it is 
contingent or FALSE you made an error.
The truth-table method you can used by calculating by hand.
You must not calculate the values for all combinations of $F$, $P$ and $S$ to see that your derivation is false. e.g.for  $P=0$, $S=0$, $F=0$ the first expression is $0$ and the second is $1$, so you have an error.
A: Step 1 and 2 look the same to me.  From step 2 to step 3 is not correct.  $A \rightarrow B$ is only false when $A$ is true and $B$ is false.  Your transition from step 3 to step 4 is the right type for step 2 to step 3.
A: $\fbox{Original}$ $(\overline{F} \land P) \rightarrow S$ (i've replaced your negation style)
$\fbox{Your attempt to negate it}$


*

*$\overline{((\overline{F} ∧ P)→S)}$ $\color{green}\checkmark$ (hyp)

*$\overline{((\overline{F} ∧ P)→\overline{S})}$  $\color{red}\chi$ (doesn't follow from 1)

*$\overline{(\overline{F} ∧ P)} → \overline{(\overline{S})}$ $\color{green}\checkmark$ (from 2)

*$\overline{(\overline{F} ∧ P)} ∧ S$ $\color{red}\chi$ (doesn't follow from 3)

*$(\overline{(\overline{F})} ∨ \overline{P}) ∧ S$ $\color{green}\checkmark$ (from 4)

*$(F ∨ \overline{P}) ∧ S$ $\color{green}\checkmark$ (from 5)
So, it fails for those two reasons. You explicitly asked not to give you the solution, so I won't.
