Let S = Steve is happy, G = George is happy
$1. (S \lor G) \wedge (\lnot S \lor (\lnot G)) $
One of Steve or George is happy and other is sad
$2. [S \lor (G \wedge (\lnot S)] \lor (\lnot G)$
George is happy when Steve is sad or George is sad
$3. S \lor [G \wedge (\lnot S \lor \lnot G) ]$
Steve is sad when George is happy or Steve is happy
Can anyone cross verify my solutions.Thank you
On
Your answer is correct.
\begin{align} [∨(∧¬)]∨¬ & \equiv [(∨)∧(∨¬)]∨¬ \\ & \equiv [(∨)∧\top]∨¬ \\ & \equiv (∨)∨¬ \\ & \equiv ∨(∨¬) \\ & \equiv ∨\top \\ & \equiv \top \end{align}
Therefore, the proposition is a tautology, and
Steve and George can have any mood.
\begin{align} S∨[G∧(¬S∨¬G)] & \equiv S∨[(G∧¬S)∨(G∧¬G))] \\ &\equiv S∨[(G∧¬S)∨\bot)] \\ &\equiv S∨(G∧¬S) \\ &\equiv (S∨G)∧(S∨¬S) \\ &\equiv (S∨G)∧(S∨¬S) \\ &\equiv (S∨G)∧\top \\ &\equiv S∨G \\ &\end{align}
Either Steve or George is happy, could be both.
As we define that not happy if and only if sad, we have $1.$ traslated correctly. But there are some ambiguity for the second one, for example
$$(\neg S\to G) \lor \neg G\tag{1}$$
$$(\neg S\lor\neg G)\to G\tag{2}$$
The second expression $[S∨(G∧(¬S)]∨(¬G)$ is actually equivalent to $(1)$ which says "(George is happy when Steve is sad) or George is sad", but not $(2)$. Consider a counter example $G=S=0$, that $$(¬S→G)∨¬G=(¬S∨¬G)→G$$ $$(¬0→0)∨¬0=(¬0∨¬0)→0$$ $$0∨1=1→0$$ $$1=0$$ This proves they are not the same when George and Steve are both sad.
Similarly for the third one
$$(G\to S)\lor S\tag{3}$$
$$(G\lor S)\to S\tag{4}$$ However, interestingly this time $(3)$ and $(4)$ are equivalent since \begin{align} (G\to S)\lor S\equiv&\neg G\lor S\lor S\\ \equiv&\neg G\lor S\\ \equiv&\neg G\lor S\land \top\\ \equiv&(\neg G\lor S)\land(\neg S\lor S)\\ \equiv&(\neg G\land\neg S)\lor S\\ \equiv&(G\lor S)\to S \end{align} And the expression $(S∨(G∧(¬S∨¬G)))$ is equivalent to $\neg S\to G$ which says that