Im having quite a bit of trouble understanding the Domination and Contraposition Laws in the instance below. I just do not see how the Domination Law, $\rho \wedge \mathrm{F} \leftrightarrow \mathrm{F}$ or $\rho \vee \mathrm{T} \leftrightarrow \mathrm{T}$, works at all on line 5. I also am confused on how the Law of Contraposition is working in line 6. Please help me understand this better. Thank you!
1 Answer
$\begingroup$Line 3 states $\sim\!s$, i.e. that $s$ is false. Then use domination rule with $\rho=r$, and regard $s$ as $F$: $$s\land r \equiv F$$ which is just the 5th line [$\sim\!(s\land r)$], and it came only from line 3.
The 6th line is derived from the 4th and 5th line by the law of Contraposition: if something of the form $A\to B$ is proven and $\sim\!B$ is also known, then $\sim\!A$ follows. Take $A=(\sim\!p\,\lor\sim\!q)$ and $B=(s\land r)$.
$\endgroup$ 2
