TITLE: Logical proof without truth table
QUESTION [1 upvotes]: 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?

REPLY [2 votes]: Your proof by logical algebra is correct.
Another method is that to show $\phi\land\psi\to\rho$ is a tautology it suffices to show that $\phi\land\psi\land\lnot\rho$ is a contradiction. This is called resolution.
$$\begin{align}P(w,s,b) &= ((w \wedge s \rightarrow b)\wedge (\neg w \rightarrow \neg s)\wedge s) \rightarrow b\\&=((\lnot w\lor\lnot s\lor b)\land(w\lor\lnot s)\land s)\to b \\\lnot P(w,s,b)&= (\lnot w\lor b\lor\lnot s)\land(w\lor\lnot s)\land s\land\lnot b\\&=(\lnot w\lor b)\land w\land s\land\lnot b\\&=\lnot w\land w\land s\land\lnot b\\&=\bot\\P(w,s,b)&=\top \end{align}$$
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