Problem
Proof following as tautology without using truth table.
$$ P(w,s,b) = ((w \wedge s \rightarrow b)\wedge (\neg w \rightarrow \neg s)\wedge (s) \rightarrow b $$
Attempt to proof
$$ (w \wedge s \rightarrow b) \iff \neg(w \wedge s) \vee b \iff (\neg w \vee \neg s) \vee b $$
$$ (\neg w \rightarrow \neg s) \iff (w \vee \neg s) $$
Now combining these we have
$$ (\neg w \vee \neg s \vee b)\wedge (s \wedge w) \to b $$ $$ \iff \neg ((\neg w \vee \neg s \vee b) \wedge (s \wedge w))\vee b $$ $$ \iff (w \wedge s \wedge \neg b) \vee (\neg s \vee \neg w \vee b) $$
Now not quite sure how should i proceed from here but if
let $$ A = (w \wedge s \wedge \neg b) $$ then we have $$ \neg A = \neg(w \wedge s \wedge \neg b) \iff \neg A = (\neg w \vee \neg s \vee b) $$
Meaning $$ (w \wedge s \wedge \neg b) \vee (\neg s \vee \neg w \vee b) \iff A \vee \neg A $$
and $A \vee \neg A$ is clearly tautology.
Is my proof correct?