Simplifying propositional logic using 'laws of logic'

Question

I am trying to simplify this: $$(h∨w)∧(\neg h∨w)∧\neg h$$ It needs to be simplified to: $$w∧\neg h$$ I have tried multiple times but cant seem the get it. I understand the laws that need to be used but I'm not getting it to work. I think the issue is with the first step, I have no idea what to do for that. Thanks

Answer 1

Note that the "not $h$" ($\neg h$) on the right hand side does not change, so we will ignore it.

To do it algebraically, there are distributive laws for compositions of "and" and "or", and the one that we want is $$ P \vee (Q \wedge R) \equiv (P \vee Q) \wedge (P \vee R). $$ Clearly your $P$ is $w$, so that by working backwards you have $$ (h \vee w) \wedge (\neg h \vee w) \equiv w \vee (h \wedge \neg h). $$ But you cannot have $(h \wedge \neg h)$ because this means that $h$ is both true and false at the same time, so that $w$ is the only true value between $w$ and $h \wedge \neg h$. Thus $w \vee (h \wedge \neg h) \equiv w$ and you're done.

You could also use the following truth table:

\begin{array}{ccc|c|c|c} h & w & \neg h & h \vee w & \neg h \vee w & (h \vee w) \wedge (\neg h \vee w) \\ \hline T & T & F & T & T & T \\ T & F & F & T & F & F \\ F & T & T & T & T & T \\ F & F & T & F & T & F \\ \end{array}

The final column is identical to the $w$ column, so that $$ (h \vee w) \wedge (\neg h \vee w) \equiv w. $$

Answer 2

If $\it{(h \lor w) \land (\neg h \lor w) \land \neg h}$, then we know $\it{\neg h}$ must be true for the statement to be true.

Since $\it{\neg h}$ must be true, then $\it{h}$ must be false, meaning if $\it{h \lor w}$ is true, then $\it{w}$ must be true.

Therefore we have $\it{w \land (\neg h \lor w) \land \neg h}$, where $\it{(\neg h \lor w)}$ is clearly redundant, giving us $\it{w \land \neg h}$.

Hope that helps.

Answer 3

Look at the first two objects in the parenthesis. I struggled with this step too. $$ (h\lor w)\land(\lnot h\lor w) $$

In this case, $\lnot h$ and $h$ both have an "or $w$" with them. You can use the Distributive Law, in reverse so that you end up with $$ w\lor (h\land \lnot h)\lor \lnot h $$

According to the Complement Law, $(h\land\lnot h)=F$. Your entire statement would then look like $$ w\lor F\lor \lnot h $$

According to the Domination law, anything "or-ed" with F is itself. Thus, you get exactly what you were looking for! $$ w\lor \lnot h $$

Hope this helped!