Semantics: The meaning of a phrase
//java
int x = 2 + 3;
(* ocaml *)
let x = 2 + 3;;
# ruby/python
x = 2 + 3
// go
x := 2 + 3
// javascript
var x = 2 + 3;
5 Idioms, 1 'semantic'
OpSem ultimately creates a proof of correctness or properties
Syntax for this class:
Goal: create a definitional interpreter
We will create rules for how an ocaml program will execute
Suppose our language is small: only numbers
(* Grammar *)
E -> n
An interpreter needs a rule of what an expression returns
An interpreter needs a rule of what an expression returns
\(e \Rightarrow v\)
Let us add addition to our language
(* Grammar *)
E -> n| E + E
\(e \Rightarrow v\)
(* Grammar *)
E -> n| E + E
\(e \Rightarrow v\)
This is an argument structure
\[\begin{array}{rl} & e_1 \Rightarrow n_1\\ & e_2 \Rightarrow n_2\\ & n_1 + n_2 = n_3\\\hline \therefore & e_1 + e_2 \Rightarrow n_3\\ \end{array}\]
Rules of inference:
\[\frac{H_1 ... H_n}{C}\]
Syntax for the class:
\[\frac{H_1 ... H_n}{C}\]
\[\frac{e1 \Rightarrow n1\qquad e2 \Rightarrow n2\qquad n3\ \text{is}\ n1+n2}{e1+e2 \Rightarrow n3}\]
Syntax for the class:
\[\frac{H_1 ... H_n}{C}\]
\[\frac{e1 \Rightarrow n1\qquad e2 \Rightarrow n2\qquad n3\ \text{is}\ n1+n2}{e1+e2 \Rightarrow n3}\]
Note: n3 is n1+n2 is in the meta language
Meta Language: the language used to describe the target language
Target Language: the language we are describing
(* Grammar *)
E -> n|E + E
Suppose \(e\) is a number \(n\):
\[\frac{}{n \Rightarrow n}\]
Suppose \(e\) is a an expression of \(e1 + e2\):
\[\frac{e1 \Rightarrow n1\qquad e2 \Rightarrow n2\qquad n3\ \text{is}\ n1+n2}{e1+e2 \Rightarrow n3}\]
Let's add more to the language
(* Grammar *)
E -> x|n|E + E|let x = E in E
We need an environment \(A\) to store variables and their values
(* Grammar *)
E -> x|n|E + E|let x = E in E
Suppose \(e\) is \(x\):
\[\frac{A(x) = v}{A; x \Rightarrow v}\]
(* Grammar *)
E -> x|n|E + E|let x = E in E
Suppose \(e\) is \(x\):
\[\frac{A(x) = v}{A; x \Rightarrow v}\]
Suppose \(e\) is \(\text{let }x = e_1\ \text{in}\ e_2\):
\[\frac{A;e_1\Rightarrow v_1 \qquad A,x:v_1;e_2\Rightarrow v_2}{A;\text{let }x = e_1\ \text{in}\ e_2 \Rightarrow v2}\]
(* Grammar *)
E -> x|n|E + E|let x = E in E
Putting it all together:
Suppose \(e\) is a number \(n\):
\[\frac{}{A;n \Rightarrow n}\]
Suppose \(e\) is a an expression of \(e1 + e2\):
\[\frac{A;e1 \Rightarrow n1\qquad A;e2 \Rightarrow n2\qquad A;n3\ \text{is}\ n1+n2}{A;e1+e2 \Rightarrow n3}\]
Suppose \(e\) is \(x\):
\[\frac{A(x) = v}{A; x \Rightarrow v}\]
Suppose \(e\) is \(\text{let }x = e_1\ \text{in}\ e_2\):
\[\frac{A;e_1\Rightarrow v_1 \qquad A,x:v_1;e_2\Rightarrow v_2}{A;\text{let }x = e_1\ \text{in}\ e_2 \Rightarrow v2}\]
Putting it all together:
Suppose \(e\) is a number \(n\):
\[\frac{}{A;n \Rightarrow n}\]
Suppose \(e\) is a an expression of \(e1 + e2\):
\[\frac{A;e1 \Rightarrow n1\qquad A;e2 \Rightarrow n2\qquad A;n3\ \text{is}\ n1+n2}{A;e1+e2 \Rightarrow n3}\]
Suppose \(e\) is \(x\):
\[\frac{A(x) = v}{A; x \Rightarrow v}\]
Suppose \(e\) is \(\text{let }x = e_1\ \text{in}\ e_2\):
\[\frac{A;e_1\Rightarrow v_1 \qquad A,x:v_1;e_2\Rightarrow v_2}{A;\text{let }x = e_1\ \text{in}\ e_2 \Rightarrow v2}\]
Time to derive/create proofs
Suppose \(e\) is a number \(n\):
\[\frac{}{A;n \Rightarrow n}\]
Suppose \(e\) is a an expression of \(e1 + e2\):
\[\frac{A;e1 \Rightarrow n1\qquad A;e2 \Rightarrow n2\qquad A;n3\ \text{is}\ n1+n2}{A;e1+e2 \Rightarrow n3}\]
Suppose \(e\) is \(x\):
\[\frac{A(x) = v}{A; x \Rightarrow v}\]
Suppose \(e\) is \(\text{let }x = e_1\ \text{in}\ e_2\):
\[\frac{A;e_1\Rightarrow v_1 \qquad A,x:v_1;e_2\Rightarrow v_2}{A;\text{let }x = e_1\ \text{in}\ e_2 \Rightarrow v2}\]
If these are the rules of our language: prove that 2+4 is both valid in the language and evaluates to 6
If these are the rules of our language: prove that 2+4 is both valid in the language and evaluates to 6
2+4 is a an expression of \(e1 + e2\):
\[\frac{A;e1 \Rightarrow n1\qquad A;e2 \Rightarrow n2\qquad A;n3\ \text{is}\ n1+n2}{A;e1+e2 \Rightarrow n3}\]
\[\cfrac{\cfrac{}{A;2 \Rightarrow 2}\qquad \cfrac{}{A;4 \Rightarrow 4}\qquad A;6\ \text{is}\ 2+4}{A;2+4 \Rightarrow 6}\]
If these are the rules of our language: prove that 2+4 is both valid in the language and evaluates to 6
2+4 is a an expression of \(e1 + e2\):
\[\frac{A;e1 \Rightarrow n1\qquad A;e2 \Rightarrow n2\qquad A;n3\ \text{is}\ n1+n2}{A;e1+e2 \Rightarrow n3}\]
\[\frac{\cfrac{}{A;2 \Rightarrow 2}\qquad \cfrac{}{A;4 \Rightarrow 4}\qquad A;6\ \text{is}\ 2+4}{A;2+4 \Rightarrow 6}\]
Now prove that let x = 3 in x + 4 is both valid in the language and evaluates to 7
Now prove that let x = 3 in x + 4 is both valid in the language and evaluates to 7
\[\frac{}{A;\text{let }x = 3\ \text{in}\ x+4 \Rightarrow 7}\]
Suppose \(e\) is a number \(n\):
\[\frac{}{A;n \Rightarrow n}\]
Suppose \(e\) is a an expression of \(e1 + e2\):
\[\frac{A;e1 \Rightarrow n1\qquad A;e2 \Rightarrow n2\qquad A;n3\ \text{is}\ n1+n2}{A;e1+e2 \Rightarrow n3}\]
Suppose \(e\) is \(x\):
\[\frac{A(x) = v}{A; x \Rightarrow v}\]
Suppose \(e\) is \(\text{let }x = e_1\ \text{in}\ e_2\):
\[\frac{A;e_1\Rightarrow v_1 \qquad A,x:v_1;e_2\Rightarrow v_2}{A;\text{let }x = e_1\ \text{in}\ e_2 \Rightarrow v2}\]
Now prove that let x = 3 in x + 4 is both valid in the language and evaluates to 7
\[\cfrac{\cfrac{}{A;3\Rightarrow 3}\qquad \cfrac{\cfrac{A,x:3(x)=3}{A,x:3;x\Rightarrow 3}\qquad\cfrac{}{A,x:3;4\Rightarrow 4}\qquad 7\text{ is }3+4}{A,x:3;x+4\Rightarrow 7}}{A;\text{let }x = 3\ \text{in}\ x+4 \Rightarrow 7}\]
As our language gets more complicated, the more rules we need to have
(* Grammar *)
E -> x|n|E + E|let x = E in E
|true|false|eq0 E
\[\frac{}{A;true \Rightarrow true}\]
\[\frac{}{A;false \Rightarrow false}\]
\[\frac{A;e \Rightarrow 0}{A;\text{eq0 } e \Rightarrow true}\]
\[\frac{A;e \Rightarrow v\qquad v \neq 0}{A;\text{eq0 } e \Rightarrow false}\]
e -> x
|λx.e
|e e
\[\cfrac{}{A;x \Rightarrow x}\]
\[\cfrac{A;e \Rightarrow e'}{A;\lambda x.e \Rightarrow \lambda x.e'}\]
\[\cfrac{A;e_1 \Rightarrow e'}{A;(e_1\ e_2) \Rightarrow (e' e_2)}\]
\[\cfrac{A;e_2 \Rightarrow e'}{A;(e_1\ e_2) \Rightarrow (e_1\ e')}\]
\[\cfrac{A;e_2 \Rightarrow e'\qquad A,x:e';e_1 \Rightarrow e''}{A;((\lambda x.e_1)\ e_2) \Rightarrow e''}\]
\[\cfrac{}{A;x \Rightarrow x}\]
\[\cfrac{A;e \Rightarrow e'}{A;\lambda x.e \Rightarrow \lambda x.e'}\]
\[\cfrac{A;e_1 \Rightarrow e'}{A;(e_1\ e_2) \Rightarrow (e' e_2)}\]
\[\cfrac{A;e_2 \Rightarrow e'}{A;(e_1\ e_2) \Rightarrow (e_1\ e')}\]
\[\cfrac{A;e_2 \Rightarrow e'\qquad A,x:e';e_1 \Rightarrow e''}{A;((\lambda x.e_1)\ e_2) \Rightarrow e''}\]
Which are Lazy? Which are Eager?
Note: level of abstractness can be Arbitrary
Note: level of abstractness can be Arbitrary
def f(x)
x = x * x
return x + 1
\[\cfrac{}{A;f(e) \Downarrow v}\]
\[\cfrac{A;e \Rightarrow n \qquad v\ is\ 1+(n*n)}{A;f(e) \Rightarrow v}\]
Like step into vs step over from GDB
Enter: (a subset of) LOLCODE
3 BTW Single number
SUM OF 3 AN 3 BTW 6
I HAS A var ITZ 5 BTW var = 5
SUM OF var AN 3 BTW 8
3 BTW Single number
SUM OF 3 AN 3 BTW 6
I HAS A var ITZ 5 BTW var = 5
SUM OF var AN 3 BTW 8
\[\frac{}{A;n \Rightarrow n}\]
\[\frac{A;e_1 \Rightarrow n_1\qquad A;e_2 \Rightarrow n_2\qquad A;n_3\ \text{is}\ n_1+n_2}{A;SUM\ OF\ e_1\ AN\ e_2 \Rightarrow n_3}\]
\[\frac{A;y \Rightarrow v_1\qquad A,x:y;e2 \Rightarrow v_2}{A;I\ HAS\ A\ x\ ITZ\ y\ \text{\\}n\ e_2 \Rightarrow v_2}\]
\[\frac{A(x) = v}{A;x \Rightarrow v}\]