CW = A`.B.C`.D` + A.B`.C`.D` + A.B`.C.D` + A.B.C`.D + A.B.C.D`
CW = A`.B.C`.D` + A.B`.D`(C`+C) + A.B(C`.D + C.D`)
CW = A`.B.C`.D` + A.B`.D`(1) + A.B(C`.D +D`.C)
CW = A`.B.C`.D` + A.B`.D` + A.B(C`.1.C)
CW = A`.B.C`.D` + A.B`.D` + A.B(1)
CW = A`.B.C`.D` + A.B`.D` + A.B
CW = A.B + A`.B.C`.D` + A.B`.D`
CW = B(A+A`.C`.D`) + A.B`.D`
CW = B(1.C`.D`) + A.B`.D`
CW = B(1) + A.B`.D`
CW = B + A.B`.D`
CW = B + B`.A.D`
CW = 1.A.D`
CW = 1
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Eric Wofsey
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No first step is wrong, ABC`D + ABCD` does not equal AB(C`D + CD`). – StainlessSteelRat Dec 23 '15 at 21:52
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$$\overline A\ B\ \overline C \ \overline D + A\ \overline B \ \overline C\ \overline D + A\overline B C\overline D + AB\overline C D + ABC\overline D$$
Look for common terms. To eliminate a term, the other 3 must be the same. Duplicate terms as required.
$$\overline A\ B\ \overline C \ \overline D + (A\ \overline B \ \overline C\ \overline D + A\overline B C\overline D) + AB\overline C D + (ABC\overline D + A\overline B C\overline D)$$ $$\overline A\ B\ \overline C \ \overline D + A\ \overline B \overline D (\overline C\ + C) + AB\overline C D + AC\overline D(B + \overline B)$$ $$\overline A\ B\ \overline C \ \overline D + A\ \overline B\ \overline D + AB\overline C D + AC\overline D$$
Best you can do.
