If $t \leq s$ and $t \neq s$, why $t' \land s \neq 0$

Question

In boolean algebra, why if $t \leq s$ and $t \neq s$, why $t' \land s \neq 0$, where ' is the complement. I see it if I use that all boolean algebras are $\{0,1\}^n$ but I want to have a proof using only the axioms.

Answer 1

Note that $$s=1\wedge s=(t\vee t')\wedge s=(t\wedge s)\vee(t'\wedge s).$$ Now $t\leq s$ means $t\wedge s=t$. So, if $t'\wedge s$ were $0$, we would get $$s=t\vee 0=t$$ which is a contradiction.