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Can you please help me solve this ? This exercise says that we have the following statements:

$$\lnot a \rightarrow b\tag{1}$$ $$\lnot a \Leftrightarrow c\tag{2}$$ $$\lnot b \rightarrow d\tag {3}$$ $$\lnot a \rightarrow d\tag {4}$$ $$ \lnot d\tag{5}$$

All this statements are all simultaneously right, so I must find the values (true or false) for a,b,c,d. The problem is , I've already done three times the simplifications using: these laws(Section Laws ), but still had the same result $a \land \lnot d \land b$, and every time $c$ is disappearing. Can you please check if it's possible to get a result consisting of all a,b,c and d ? Thanks.

amWhy
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wonderingdev
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2 Answers2

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$a \land b \land \lnot d$ is a good start. With respect to $c$, one way to see that we must have $\land \lnot c$ is from the premise $$\lnot a \iff c$$

(Remember, $p \iff q$ means both $p, q$ are false, or both $p, q$ are true.)

Since we have $a$ (true), $\lnot a$ (false), then by the biconditional $\lnot a \iff c$, we know that $c$ must also be false (has the same truth value as $\lnot a$), hence $\lnot c$ true.

With your own results, that gives us $$a\land b \land \lnot c \land \lnot d$$

amWhy
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I think I have a solution.

From (5), we get $\lnot d$.

Now consider (3) or (4). If $\lnot a \Rightarrow d$, since we have $\lnot d$, we must have $\lnot(\lnot a)$, or we have a contradiction. Thus $a$, and similarly $b$.

Finally, (2) has $\lnot a \Leftrightarrow c$, and if $c$ were true, then $c \Rightarrow \lnot a$ would be a contradiction. Thus $\lnot c$.

It follows that the solution is $a \land b \land (\lnot c) \land (\lnot d)$.

alexwlchan
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