Axiomatic definitions of sublanguages of Loglan'82: Różnice pomiędzy wersjami

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Guess what will print this program.
 
Guess what will print this program.
  
One need not to guess. We have much better tool.  
+
One need not to guess. We have much better tool. <br />
 
<big>Lemma</big><br />
 
<big>Lemma</big><br />
<math>(c_1=a \land c_2=b)\Rightarrow [c_3:=c_1; c_1:=c_2; c_2:=c_3](c_1=b \land c_2=a) </math>
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<math>(c_1=a \land c_2=b)\Rightarrow [c_3:=c_1; c_1:=c_2; c_2:=c_3](c_1=b \land c_2=a) </math><br />
I.e. ''the execution of three instructions causes swap of the values of variables'' <math> c_1, c_2 </math>.
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I.e. ''the execution of three instructions causes swap of the values of variables'' <math> c_1, c_2 </math>.<br />
 
'''Proof''' uses two axioms of algorithmic logic: <br />
 
'''Proof''' uses two axioms of algorithmic logic: <br />
 
<math>\begin{equation}\tag{Ax:=}[x:=\tau]\alpha \Leftrightarrow \alpha(x/\tau) \end{equation} </math> <br />
 
<math>\begin{equation}\tag{Ax:=}[x:=\tau]\alpha \Leftrightarrow \alpha(x/\tau) \end{equation} </math> <br />
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<math>\begin{equation}\tag{Ax ;}[K; M]\alpha \Leftrightarrow [K][M]\alpha  \end{equation} </math><br />
 
<math>\begin{equation}\tag{Ax ;}[K; M]\alpha \Leftrightarrow [K][M]\alpha  \end{equation} </math><br />
 
and the inference rule  
 
and the inference rule  
<math>\begin{equation}\tag{R2}\dfrac{\alpha \Rightarrow \beta}{M\alpha \Rightarrow M\beta} \end{equation} </math>
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<math>\begin{equation}\tag{R2}\dfrac{\alpha \Rightarrow \beta}{M\alpha \Rightarrow M\beta}   \end{equation} </math>
 
Now observe that the following formulas are equivalent<br />
 
Now observe that the following formulas are equivalent<br />
 
<math>\begin{align*}{}
 
<math>\begin{align*}{}
 
&(c_1 =a \land c_2=b)&            & \\
 
&(c_1 =a \land c_2=b)&            & \\
\Rightarrow &(c_1 =a \land c_2=b \land c1=a)&            & \mbox{by propositional calculus} \\
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\Leftrightarrow &(c_1 =a \land c_2=b \land c1=a)&            & \mbox{by propositional calculus} \\
\Rightarrow & [c_3:=c_1](c_3=a \land c_2 =b \land c_3=a) &      & \mbox{by Ax :=} \\
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\Leftrightarrow & [c_3:=c_1](c_3=a \land c_2 =b \land c_3=a) &      & \mbox{by Ax :=} \\
\Rightarrow &  [c_3:=c_1][c_1:=c_2](c_3=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax := and R2} \\
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\Leftrightarrow &  [c_3:=c_1][c_1:=c_2](c_3=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax := and R2} \\
\Rightarrow &  [c_3:=c_1; c_1:=c_2](c_3=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax ;} \\
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\Leftrightarrow &  [c_3:=c_1; c_1:=c_2](c_3=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax ;} \\
\Rightarrow &  [c_3:=c_1; c_1:=c_2][c_2 :=c_3](c_2=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax := and R2}\\
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\Leftrightarrow &  [c_3:=c_1; c_1:=c_2][c_2 :=c_3](c_2=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax := and R2}\\
\Rightarrow &  [c_3:=c_1; c_1:=c_2; c_2 :=c_3](c_2=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax ;} \\
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\Leftrightarrow &  [c_3:=c_1; c_1:=c_2; c_2 :=c_3](c_2=a \land c_1 =b \land c_3=a) &      & \mbox{by Ax ;} \\
 
\end{align*}  </math>
 
\end{align*}  </math>
  

Aktualna wersja na dzień 15:21, 27 lis 2014

On these pages we shall attempt to develop a sequence of algorithmic theories that correspond to a sequence of sublanguages of of Loglan'82 language.


Here we shall present an increasing sequence of sublanguages [math]\mathcal{L}_0\subset\mathcal{L}_1\subset\mathcal{L}_2\subset \mathcal{L}_3\subset\mathcal{L}_4\subset \dots[/math]Loglan'82. For each language [math]\mathcal{L}_i[/math] we shall present its grammar and some axioms and inference rules that define its semantics.

Program

Program in Loglan'82 has the following structure

Program
definiendum definiens
program program <name>;
<declarations>

begin

<instructions>

end

where name is any identifier, i.e. a finite sequence of letters and digits beginning with a letter.

The declarations and instructions are finite sequences of declarations and instructions respectively, empty squence included.

This allow us to define the first sublanguage [math]\mathcal{L}_0[/math] of Loglan'82.

[math]\mathcal{L}_0 \stackrel{df}{=}\{p\in \mathcal{A}^*: p=\textbf{program}\ \textit{id}; \textbf{begin end} \}[/math]

Programs of the language [math]\mathcal{L}_0[/math] are empty programs, they posses just a name. Their list of instructions as well as list of declarations are empty. The effect of execution of such program is do nothing.

Now we shall define a new language [math]\mathcal{L}_{0.1}[/math]. First, we say that any expression of the form:
writeln;
write(integer);
write("here your text");
is an output instruction.

[math]\mathcal{L}_{0.1} \stackrel{df}{=}\{p\in \mathcal{A}^*: p=\textbf{program} \textit{ id}; \textbf{begin} \lt\textit{output instructions}\gt \textbf{end} \}[/math]

Example 0.1
program print;

begin

write("hallo world!"); writeln;
write("Today is: May"); write(22); write(2014); writeln

end

Primitive types

Characters

The type char is a finite set of characters. No operation is defined on charcters. However one can compare two characters.

'a' = 'a'

'a' =/= 'c'

The language [math]\mathcal{L}_{1.1} [/math] allows to declare variables of type char and to assign values to the variables. Assignments

Example
program characters;

var c1,c2,c3:char

begin

c1:='p';
c2:='d';
writeln(c1,c2);
c3:=c1;
c1:=c2;
c2:=c3;
writeln(c1,c2)

end

Exercise
Guess what will print this program.

One need not to guess. We have much better tool.
Lemma
[math](c_1=a \land c_2=b)\Rightarrow [c_3:=c_1; c_1:=c_2; c_2:=c_3](c_1=b \land c_2=a) [/math]
I.e. the execution of three instructions causes swap of the values of variables [math] c_1, c_2 [/math].
Proof uses two axioms of algorithmic logic:
[math]\begin{equation}\tag{Ax:=}[x:=\tau]\alpha \Leftrightarrow \alpha(x/\tau) \end{equation} [/math]
this axiom of assignment instruction reads: formula [math]\alpha(x/\tau)[/math] holds iff after execution of instruction [math]x:=\tau[/math] formula [math]\alpha[/math] holds.
[math]\begin{equation}\tag{Ax ;}[K; M]\alpha \Leftrightarrow [K][M]\alpha \end{equation} [/math]
and the inference rule [math]\begin{equation}\tag{R2}\dfrac{\alpha \Rightarrow \beta}{M\alpha \Rightarrow M\beta} \end{equation} [/math] Now observe that the following formulas are equivalent
[math]\begin{align*}{} &(c_1 =a \land c_2=b)& & \\ \Leftrightarrow &(c_1 =a \land c_2=b \land c1=a)& & \mbox{by propositional calculus} \\ \Leftrightarrow & [c_3:=c_1](c_3=a \land c_2 =b \land c_3=a) & & \mbox{by Ax :=} \\ \Leftrightarrow & [c_3:=c_1][c_1:=c_2](c_3=a \land c_1 =b \land c_3=a) & & \mbox{by Ax := and R2} \\ \Leftrightarrow & [c_3:=c_1; c_1:=c_2](c_3=a \land c_1 =b \land c_3=a) & & \mbox{by Ax ;} \\ \Leftrightarrow & [c_3:=c_1; c_1:=c_2][c_2 :=c_3](c_2=a \land c_1 =b \land c_3=a) & & \mbox{by Ax := and R2}\\ \Leftrightarrow & [c_3:=c_1; c_1:=c_2; c_2 :=c_3](c_2=a \land c_1 =b \land c_3=a) & & \mbox{by Ax ;} \\ \end{align*} [/math]

Boolean

Syntax

Axioms


Example
program Boole;

var a, b, c, b1, b2:Boolean

begin

b1:=a or b;
b2:=b1 and c;
writeln(b1, b2)

end

Declarations of variables. Assignment instructions

Example

program P2 ;

var x,y: integer;

begin

y:=74;
x:= y+ 8;

end

Grammar
Context free grammar

Well formed expressions

Axiom

[math]\{x:=\tau\}( \alpha ) \Leftrightarrow \alpha(x/\tau)[/math]