Boolean algebra laws Can someone explain to me why in Boolean algebra

$$ f(x,y,z,t)=z+x'y+xy'+xt'+yt'
                =z+x'y+xy'+xt'$$

I have no clue why u can just leave out the last term, is it due to some absorption law.
Ps: $x'$ is the complement of $x$.
 A: The consensus theorem helps reduce your expression. It is a very important method in digital design for reducing boolean expressions. I'll provide an algebraic proof below:
\begin{align}
x^{\prime} y + x t^{\prime} + y t^{\prime} &= x^{\prime} y + x t^{\prime} + y t^{\prime} (x + x^{\prime}) \\
&= x^{\prime} y + x t^{\prime} + x y t^{\prime} + x^{\prime} y t^{\prime} \\
&= x^{\prime} y + x^{\prime} y t^{\prime} + x t^{\prime} + x y t^{\prime} \\
&= x^{\prime} y (1 + t^{\prime}) + x t^{\prime} (1 + y) \\
&= x^{\prime} y + x t^{\prime}.
\end{align}
A: For a Boolean algebra B let $B^*= \{ p^* |p \in B \}$ where $p^*= \{ q \in B |qp=q \}$ for $p \in B$. Then $(pq)^*=p^* \cap q^*$ and $(p+q)^*=p^* \cup q^*$ for $p,q \in B$,and $p'^*=B^* -p^* .$ So $B$ is isomorphic to the Boolean algebra $B^*$ ,where addition and multiplication in $B^*$ are  $ \cup$ and $ \cap$, and $x'$ in $B^*$ means $B-x$ .So, if you consider $f$ as if plus and times are $\cup$ and $\cap$,and $p'$ is the complement of $p$, it works out easily .
