$S,C,D,O$ are statements.
Then $((\neg S\rightarrow C)\wedge(C\rightarrow\neg D)\wedge(D\vee O)\wedge \neg O)\rightarrow S$ is a tautology.
This can be checked by a truth table or by the following.
(1) $\neg S\rightarrow C$
(2) $C\rightarrow\neg D$
(3) $D\vee O$
(4) $\neg O$
((1),(2),(3),(4) are premises.)
(5) $D$ ((3),(4)$\implies$(5))
(6) $\neg C$ ((2),(5)$\implies$(6))
(7) $S$ ((1),(6)$\implies$(7))
I don't understand why the above process, (1)-(7) verifies $((\neg S\rightarrow C)\wedge(C\rightarrow\neg D)\wedge(D\vee O)\wedge \neg O)\rightarrow S$ is a tautology.
Best Answer
If $\varphi$ and $\psi$ are formulas, then $\varphi \to \psi$ is a tautology iff $\varphi \vdash \psi$. Here, $\varphi = (\neg S\rightarrow C)\wedge(C\rightarrow\neg D)\wedge(D\vee O)\wedge \neg O$ and $\psi = S$, so you have to show that $(\neg S\rightarrow C)\wedge(C\rightarrow\neg D)\wedge(D\vee O)\wedge \neg O \vdash S$ ie : $$\{\underbrace{\neg S\rightarrow C}_{(1)} , \underbrace{C\rightarrow\neg D}_{(2)} , \underbrace{D\vee O}_{(3)}, \underbrace{\neg O}_{(4)} \} \vdash S$$
What you wrote is a proof of the above, since :
$\{(3), (4)\} \vdash D$
$\{(2), D\} \vdash C$ hence $\{(2), (3), (4)\} \vdash C$
etc