Employee1(SSN, Fname, LName, PNumber, PName, Hours) 
when we made a deviation to discuss lossless decomposition ....
Let R_{1} and
R_{2} be a
decomposition of
R
If either R_{1} ∩ R_{2} → R_{1} or R_{1} ∩ R_{2} → R_{2}, then:

decompose relations losslessly into normal forms.

Employee1(SSN, Fname, LName, PNumber, PName, Hours) Key(s): (SSN, PNumber)  Employee1 is not in 2NF 
Nonkey attribute: LName Key subset: SSN SSN → LName 
Original relation: R = Employee1(SSN, FName, LName, PNumber, PName, Hours) Violation: SSN → LName 
R1(SSN, FName, LName) Key(s): (SSN)  R1 is in 2NF 
Nonkey attribute: PName Key subset: PNumber PNumber → PName 
Original relation: R2 (SSN, PNumber, PName, Hours) Violation: PNumber → PName 
R1 = (SSN, FName, LName) R21 = (PNumber, PName) R22 = (SSN, PNumber, Hours) 
R (a, b, c, d, e, f) Functional Dependencies: a → b,c d → e ad → f 

To answer this question, we must find all the keys of R.
Functional Dependencies: a → b,c d → e ad → f Necessary attributes to include in key: a, d Test for sufficiency: ad^{+} = a, d, b, c, e, f  sufficient ! 
R (a, b, c, d, e, f) Functional Dependencies: a → b,c d → e ad → f Key: (a, d) 
R (a, b, c, d, e, f) Functional Dependencies: a → b,c d → e ad → f Key: (a, d) 
Functional Dependencies: a → b,c d → e ad → f 
Functional Dependencies: a → b,c d → e ad → f 
R1 = (a, b, c) R2 = (a, d, e, f) Functional Dependencies: a → b,c d → e ad → f Key: (a, d) 
Functional Dependencies: a → b,c d → e ad → f 
Functional Dependencies: a → b,c d → e ad → f 


Can't be more objective than this....