x\'y\'z\' + x\'y\'z + x\'yz + xyz + xyz\' I need to reduce the above expression

Question

x'y'z' + x'y'z + x'yz + xyz + xyz'

I need to reduce the above expression to a minimum SOP form, this one needing 3 terms, 6 literals. There are two solutions to this problem.

Solution one: x'y' + xy + yz
Solution two: x'y' + xy + x'z

I have already reduced to solution one after a my first attempt, but I have been trying to reduce to solution two for an hour it seems. I must just be missing something. Can you help me reduce to solution two using the properties of boolean algebra, and list the properties you used?

Thank you!!

Explanation / Answer

x'y'z' + x'y'z + x'yz + xyz + xyz'

= x'y' ( z+z') + x'yz + xyz + xyz'      using identity z+z' = 1

= x'y' + x'yz + xy(z + z')         using identity z+z' = 1

= x'y' + x'yz + xy

= x'(y'+yz) + xy                just took common term out

= x' ( y'+y) (y'+z)) + xy    using identity A+BC = (A+B)(A+c)         

since y+y' = 1

= x'(y'+z) + xy

= x'y' + xy + x'z

thus ans

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