I've been having trouble simplifying this expression:
$$F = X'+ XY'Z'+ XY'ZW'+XYW + YZ'W+ X'Y'ZW'+ X'Y'Z'W'$$
I was able to reach this point:
$$F = X'+ XY'Z'+ XY'ZW'+ X'Y'ZW' +XYW + YZ'W+ X'Y'ZW'+ X'Y'Z'W'$$
by duplicating $X'Y'ZW'$
$$F = X'+ XY'Z'+ Y'ZW'+ XYW+ YZ'W+ X'Y'W'$$
Is there any way of reducing this further? If so, how?
Those computations sometimes might take some time.
We first start by finding small term in bigger terms, for example $$\boxed{X′}+XY′Z′+XY′ZW′+XYW+YZ′W+\boxed{X′}Y′ZW′+\boxed{X′}Y′Z′W′$$ Apply Idempotent law $X'=X'+X'$
Now use Commutative law to combine $X'$ with $X'Y'ZW'$ and $X'Y'Z'W'$ $\require{cancel}$ gives $$\underline{X'+X'Y'ZW'+X'+X'Y'Z'W'}+XY'Z'+XY'ZW'+XYW+YZ'W$$ Apply Absorption law twice $$\underline{X'+X'}+XY'Z'+XY'ZW'+XYW+YZ'W$$ Then use Idempotent law again to get three $X'$, and Commutative law $$X'+XY'Z'+X'+XY'ZW'+X'+XYW+YZ'W$$ So we can distribute $X'$ into $XY'Z',XY'ZW'$ and $XYW$ $$\cancel{(X'+X)}(X'+Y')(X'+Z')+\cancel{(X'+X)}(X'+Y')(X'+Z)(X'+W')+\cancel{(X'+X)}(X'+Y)(X'+W)+YZ'W$$ By Idempotent and Identity law we cancels $(X'+X)$ get $$\boxed{(X'+Y')}(X'+Z')+\boxed{(X'+Y')}(X'+Z)(X'+W')+(X'+Y)(X'+W)+YZ'W$$ Now we distriute $(X'+Y')$ out have $$(X'+Y')((X'+Z')+(X'+Z)(X'+W'))+(X'+Y)(X'+W)+YZ'W$$ Then distriute $(X'+Z')$ into $(X'+Z)(X'+W')$ gives $$(X'+Y')(\cancel{((X'+Z')+(X'+Z))}((\boxed{X'}+Z')+(\boxed{X'}+W')))+(X'+Y)(X'+W)+YZ'W$$ As we apply Commutative law have $Z+Z'$, by Identity and Domination law cancels $$(X'+Z')+(X'+Z)$$
At the same time apply Idempotent law again $X′+X′=X'$ that $$(X'+Y')(X'+Z'+W')+(X'+Y)(X'+W)+YZ'W$$ Then distribute $X'$ out of $(X′+Y′)(X′+Z′+W′)$ we get $$X'+(Y'(Z'+W'))+(X'+Y)(X'+W)+YZ'W$$ And distribut $Y'$ into $(Z′+W′)$ have $$X'+Y'Z'+Y'W'+X'+YW+YZ'W$$ Finally apply Idempotent law and Absorption law gives $$X'+Y'Z'+Y'W'+YW$$ Hence this is the simplified disjunctive normal form of the expression, which can be also verified by the Karnaugh map.