Axioms of algorithmic logic

Axioms of AL

[math]Ax_1 \qquad ((\alpha\Longrightarrow\beta )\Longrightarrow((\beta\Longrightarrow\delta)\Longrightarrow(\alpha \Longrightarrow\delta)))[/math]

\item[$Ax_2$] $(\alpha\Longrightarrow(\alpha \vee\beta))$ \item[$Ax_3$] $(\beta\Longrightarrow(\alpha\vee\beta))$%

\item[$Ax_4$] $((\alpha\Longrightarrow\delta)~\Longrightarrow((\beta \Longrightarrow\delta)~\Longrightarrow~((\alpha\vee\beta )\Longrightarrow\delta)))$ \item[$Ax_5$] $((\alpha\wedge\beta )\Longrightarrow\alpha)$ \item[$Ax_6$] $((\alpha\wedge\beta )\Longrightarrow\beta)$ \item[$Ax_7$] $((\delta\Longrightarrow\alpha )\Longrightarrow((\delta\Longrightarrow\beta)\Longrightarrow(\delta \Longrightarrow(\alpha\wedge\beta))))$ \item[$Ax_8$] $((\alpha \Longrightarrow(\beta\Longrightarrow\delta))\equiv((\alpha\wedge\beta )\Longrightarrow\delta))$ \item[$Ax_9$] $((\alpha\wedge\lnot\alpha )\Longrightarrow\beta)$ \item[$Ax_{10}$] $((\alpha\Longrightarrow (\alpha\wedge\lnot\alpha))\Longrightarrow\lnot\alpha)$ \item[$Ax_{11}$] $(\alpha\vee\lnot\alpha)$ \item[$Ax_{12}$] $((x:=\tau )true\Longrightarrow((\forall x)\alpha(x)\Longrightarrow(x:=\tau)\alpha (x)))$ \ \ \newline \qquad\qquad{\small where term }$\tau${\small\ is of the same type as the variable x} \item[$Ax_{13}$] $(\forall x)\alpha(x)\equiv\lnot (\exists x)\lnot\alpha(x)$ \item[$Ax_{14}$] $K((\exists x)\alpha (x))\equiv(\exists y)(K\alpha(x/y))$ \begin{small}for $y\notin V(K)$ \end{small} \item[$Ax_{15}$] $K(\alpha\vee\beta)\equiv((K\alpha)\vee(K\beta))$ \item[$Ax_{16}$] $K(\alpha\wedge\beta)\equiv((K\alpha)\wedge(K\beta))$% \item[$Ax_{17}$] $K(\lnot\alpha)\Longrightarrow\lnot(K\alpha)$% \item[$Ax_{18}$] $((x:=\tau)\gamma\equiv(\gamma(x/\tau)\wedge (x:=\tau)true))~\wedge((q:=\gamma\prime)\gamma\equiv\gamma(q/\gamma \prime))$ \item[$Ax_{19}$] \textbf{begin} $K;M$ \textbf{end} $\alpha \equiv K(M\alpha)$ \item[$Ax_{20}$] \textbf{if} $\gamma$ \textbf{then} $K $ \textbf{else} $M$ \textbf{fi} $\alpha\equiv((\lnot\gamma\wedge M\alpha )\vee(\gamma\wedge K\alpha))$ \item[$Ax_{21}$] \textbf{while}~ $\gamma$~\textbf{do}~$K$~\textbf{od}~$\alpha\equiv((\lnot\gamma\wedge\alpha )\vee(\gamma\wedge K($\textbf{while}~$\gamma$~\textbf{do}~$K$~\textbf{od}$% (\lnot\gamma\wedge\alpha))))$ \item[$Ax_{22}$] $\displaystyle{\bigcap K\alpha\equiv (\alpha\wedge(K\bigcap K\alpha))}$ \item[$Ax_{23}$] $\displaystyle{\bigcup K\alpha \equiv(\alpha\vee(K\bigcup K\alpha))}$ \end{trivlist}