Axioms of algorithmic logic: Różnice pomiędzy wersjami

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)\Longrightarrow\delta))</math>
 
)\Longrightarrow\delta))</math>
  
−
<span style="color: Blue"><math>Ax_9 \qquad ((\alpha\wedge\lnot\alpha
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<span style="color: Blue"><math>Ax_9 \qquad\ ((\alpha\wedge\lnot\alpha
 
)\Longrightarrow\beta)</math>
 
)\Longrightarrow\beta)</math>
  
Linia 39: Linia 39:
  
 
== Axioms of predicate calculus ==
 
== Axioms of predicate calculus ==
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\item[$Ax_{12}$] $((x:=\tau
+
<span style="color: Blue"><math>Ax_{12} \qquad ((x:=\tau
 
)true\Longrightarrow((\forall x)\alpha(x)\Longrightarrow(x:=\tau)\alpha
 
)true\Longrightarrow((\forall x)\alpha(x)\Longrightarrow(x:=\tau)\alpha
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(x)))$ \ \ \newline
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(x))) \\  
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\qquad\qquad{\small where term }$\tau${\small\ is of the same type as
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\qquad\qquad {\small where\ term\ }\tau {\ \small\ is\ of\ the\ same\ type\ as\
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the variable x}
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the\ variable\ x}</math>
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\item[$Ax_{13}$] $(\forall x)\alpha(x)\equiv\lnot
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<span style="color: Blue"><math>Ax_{13} \qquad (\forall x)\alpha(x)\equiv\lnot
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(\exists x)\lnot\alpha(x)$
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(\exists x)\lnot\alpha(x)</math>
−
\item[$Ax_{14}$] $K((\exists x)\alpha
+
 
−
(x))\equiv(\exists y)(K\alpha(x/y))$ \begin{small}for $y\notin V(K)$  \end{small}
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<span style="color: Blue"><math>Ax_{14} \qquad K((\exists x)\alpha
−
== Axioms of algorithmic logic ==                                                
+
(x))\equiv(\exists y)(K\alpha(x/y))\ \ \ \ \ \ \ \ \ \ \ \ \  \ for\ y\notin V(K) </math></span>
−
\item[$Ax_{15}$] $K(\alpha\vee\beta)\equiv((K\alpha)\vee(K\beta))$
+
 
−
\item[$Ax_{16}$] $K(\alpha\wedge\beta)\equiv((K\alpha)\wedge(K\beta))$%
+
== Axioms of algorithmic logic ==
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\item[$Ax_{17}$] $K(\lnot\alpha)\Longrightarrow\lnot(K\alpha)$%
+
<span style="color: Blue"><math>Ax_{14} \qquad K((\exists x)\alpha
−
\item[$Ax_{18}$] $((x:=\tau)\gamma\equiv(\gamma(x/\tau)\wedge
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(x))\equiv(\exists y)(K\alpha(x/y)) </math></span> for <math> y\notin V(K) </math>
−
(x:=\tau)true))~\wedge((q:=\gamma\prime)\gamma\equiv\gamma(q/\gamma
+
                                             
−
\prime))$
+
<span style="color: Blue"><math>Ax_{15} \qquad K(\alpha\vee\beta)\equiv((K\alpha)\vee(K\beta))</math></span>
−
\item[$Ax_{19}$] \textbf{begin} $K;M$ \textbf{end} $\alpha
+
 
−
\equiv K(M\alpha)$
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<span style="color: Blue"><math>Ax_{16} \qquad K(\alpha\wedge\beta)\equiv((K\alpha)\wedge(K\beta))</math></span>
−
\item[$Ax_{20}$] \textbf{if} $\gamma$ \textbf{then} $K
+
 
−
$ \textbf{else} $M$ \textbf{fi} $\alpha\equiv((\lnot\gamma\wedge M\alpha
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<span style="color: Blue"><math>Ax_{17} \qquad K(\lnot\alpha)\Longrightarrow\lnot(K\alpha)</math></span>
−
)\vee(\gamma\wedge K\alpha))$
+
 
−
\item[$Ax_{21}$] \textbf{while}~
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<span style="color: Blue"><math>Ax_{18} \qquad ((x:=\tau)\gamma\equiv(\gamma(x/\tau)\wedge
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$\gamma$~\textbf{do}~$K$~\textbf{od}~$\alpha\equiv((\lnot\gamma\wedge\alpha
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(x:=\tau)true))~\wedge((q:=\gamma^\prime)\gamma\equiv\gamma(q/\gamma^
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)\vee(\gamma\wedge K($\textbf{while}~$\gamma$~\textbf{do}~$K$~\textbf{od}$%
+
\prime))</math></span>
−
(\lnot\gamma\wedge\alpha))))$
+
 
−
\item[$Ax_{22}$] $\displaystyle{\bigcap K\alpha\equiv
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<span style="color: Blue"><math>Ax_{19} \qquad \textbf{begin } K;M \textbf{ end}\, \alpha
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(\alpha\wedge(K\bigcap K\alpha))}$
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\equiv K(M\alpha)</math></span>
−
\item[$Ax_{23}$] $\displaystyle{\bigcup K\alpha
+
 
−
\equiv(\alpha\vee(K\bigcup K\alpha))}$ </span>
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<span style="color: Blue"><math>Ax_{20} \qquad  \textbf{if}\ \gamma\ \textbf{then}\ K
 +
\ \textbf{else } M \textbf{ fi}\, \alpha\equiv((\lnot\gamma\wedge M\alpha
 +
)\vee(\gamma\wedge K\alpha))</math></span>
 +
 
 +
<span style="color: Blue"><math>Ax_{21} \qquad  \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))))</math></span>
 +
 
 +
<span style="color: Blue"><math>Ax_{22} \qquad \bigcap K\alpha\equiv
 +
(\alpha\wedge(K\bigcap K\alpha))</math></span>
 +
 
 +
<span style="color: Blue"><math>Ax_{23} \qquad \bigcup K\alpha
 +
\equiv(\alpha\vee(K\bigcup K\alpha))</math> </span>

Aktualna wersja na dzień 18:08, 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

[math]Ax_{14} \qquad K((\exists x)\alpha (x))\equiv(\exists y)(K\alpha(x/y)) [/math] for [math] y\notin V(K) [/math]

[math]Ax_{15} \qquad K(\alpha\vee\beta)\equiv((K\alpha)\vee(K\beta))[/math]

[math]Ax_{16} \qquad K(\alpha\wedge\beta)\equiv((K\alpha)\wedge(K\beta))[/math]

[math]Ax_{17} \qquad K(\lnot\alpha)\Longrightarrow\lnot(K\alpha)[/math]

[math]Ax_{18} \qquad ((x:=\tau)\gamma\equiv(\gamma(x/\tau)\wedge (x:=\tau)true))~\wedge((q:=\gamma^\prime)\gamma\equiv\gamma(q/\gamma^ \prime))[/math]

[math]Ax_{19} \qquad \textbf{begin } K;M \textbf{ end}\, \alpha \equiv K(M\alpha)[/math]

[math]Ax_{20} \qquad \textbf{if}\ \gamma\ \textbf{then}\ K \ \textbf{else } M \textbf{ fi}\, \alpha\equiv((\lnot\gamma\wedge M\alpha )\vee(\gamma\wedge K\alpha))[/math]

[math]Ax_{21} \qquad \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))))[/math]

[math]Ax_{22} \qquad \bigcap K\alpha\equiv (\alpha\wedge(K\bigcap K\alpha))[/math]

[math]Ax_{23} \qquad \bigcup K\alpha \equiv(\alpha\vee(K\bigcup K\alpha))[/math]

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