Axioms of algorithmic logic: Różnice pomiędzy wersjami

Wersja z 17:41, 19 lis 2015

The axioms are grouped in three sets

Axioms of propositional logic

We are quoting the axiom system of Helena Rasiowa and Roman Sikorski
Any formula of a scheme listed below is an axiom of algorithmic logic. The signs [math]\alpha, \beta, \delta[/math] are to be replaced by a formula. [math]Ax_1 \qquad ((\alpha\Rightarrow\beta )\Rightarrow((\beta\Rightarrow\delta)\Rightarrow(\alpha \Rightarrow\delta)))[/math]

[math]Ax_2 \qquad(\alpha\Rightarrow(\alpha \vee\beta))[/math]

[math]Ax_3 \qquad (\beta\Longrightarrow(\alpha\vee\beta))[/math]

[math]Ax_4 \qquad((\alpha\Longrightarrow\delta)~\Longrightarrow((\beta \Longrightarrow\delta)~\Longrightarrow~((\alpha\vee\beta )\Longrightarrow\delta)))[/math]

[math]Ax_5 \qquad((\alpha\wedge\beta )\Longrightarrow\alpha)[/math]

[math]Ax_6 \qquad((\alpha\wedge\beta )\Longrightarrow\beta)[/math]

[math]Ax_7 \qquad((\delta\Longrightarrow\alpha )\Longrightarrow((\delta\Longrightarrow\beta)\Longrightarrow(\delta \Longrightarrow(\alpha\wedge\beta))))[/math]

[math]Ax_8 \qquad ((\alpha \Longrightarrow(\beta\Longrightarrow\delta))\equiv((\alpha\wedge\beta )\Longrightarrow\delta))[/math]

[math]Ax_9 \qquad\ ((\alpha\wedge\lnot\alpha )\Longrightarrow\beta)[/math]

[math]Ax_{10} \qquad ((\alpha\Longrightarrow (\alpha\wedge\lnot\alpha))\Longrightarrow\lnot\alpha)[/math]

[math]Ax_{11} \qquad (\alpha\vee\lnot\alpha)[/math]

Axioms of predicate calculus

[math]Ax_{12} \qquad ((x:=\tau )true\Longrightarrow((\forall x)\alpha(x)\Longrightarrow(x:=\tau)\alpha (x))) \\ \qquad\qquad {\small where\ term\ }\tau {\ \small\ is\ of\ the\ same\ type\ as\ the\ variable\ x}[/math] [math]Ax_{13} \qquad (\forall x)\alpha(x)\equiv\lnot (\exists x)\lnot\alpha(x)[/math] [math]Ax_{14} \qquad K((\exists x)\alpha (x))\equiv(\exists y)(K\alpha(x/y))\ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ y\notin V(K) [/math]

Axioms of algorithmic logic

\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[$Ax_{23}$] $\displaystyle{\bigcup K\alpha \equiv(\alpha\vee(K\bigcup K\alpha))}$