SSL AExp.thy
Interaktion und PortierbarkeitIsabelle
section "Arithmetic and Boolean Expressions"
subsection "Arithmetic Expressions"
theory AExp imports Main begin
type_synonym vname = string type_synonym val = int type_synonym state = "vname \ val"
text_raw\<open>\snip{AExpaexpdef}{2}{1}{%\<close> datatype aexp = N int | V vname | Plus aexp aexp text_raw\<open>}%endsnip\<close>
text_raw\<open>\snip{AExpavaldef}{1}{2}{%\<close> fun aval :: "aexp \ state \ val" where "aval (N n) s = n" | "aval (V x) s = s x" | "aval (Plus a\<^sub>1 a\<^sub>2) s = aval a\<^sub>1 s + aval a\<^sub>2 s" text_raw\<open>}%endsnip\<close>
value"aval (Plus (V ''x'') (N 5)) (\x. if x = ''x'' then 7 else 0)"
text\<open>The same state more concisely:\<close> value"aval (Plus (V ''x'') (N 5)) ((\x. 0) (''x'':= 7))"
text\<open>A little syntax magic to write larger states compactly:\<close>
text\<open>\noindent
We can now write a series of updates to the function\<open>\<lambda>x. 0\<close> compactly: \<close> lemma" = (<> (a := 1)) (b := (2::int))" by (rule refl)
value"aval (Plus (V ''x'') (N 5)) <''x'' := 7>"
text\<open>In the @{term[source] "<a := b>"} syntax, variables that are not mentioned are 0 by default: \<close> value"aval (Plus (V ''x'') (N 5)) <''y'' := 7>"
text\<open>Note that this \<open><\<dots>>\<close> syntax works for any function space \<open>\<tau>\<^sub>1 \<Rightarrow> \<tau>\<^sub>2\<close> where \<open>\<tau>\<^sub>2\<close> has a \<open>0\<close>.\<close>
text_raw\<open>\snip{AExpasimpconstdef}{0}{2}{%\<close> fun asimp_const :: "aexp \ aexp" where "asimp_const (N n) = N n" | "asimp_const (V x) = V x" | "asimp_const (Plus a\<^sub>1 a\<^sub>2) =
(case (asimp_const a\<^sub>1, asimp_const a\<^sub>2) of
(N n\<^sub>1, N n\<^sub>2) \<Rightarrow> N(n\<^sub>1+n\<^sub>2) |
(b\<^sub>1,b\<^sub>2) \<Rightarrow> Plus b\<^sub>1 b\<^sub>2)" text_raw\<open>}%endsnip\<close>
theorem aval_asimp_const: "aval (asimp_const a) s = aval a s" apply(induction a) apply (auto split: aexp.split) done
text\<open>Now we also eliminate all occurrences 0 in additions. The standard
method: optimized versions of the constructors:\<close>
text_raw\<open>\snip{AExpplusdef}{0}{2}{%\<close> fun plus :: "aexp \ aexp \ aexp" where "plus (N i\<^sub>1) (N i\<^sub>2) = N(i\<^sub>1+i\<^sub>2)" | "plus (N i) a = (if i=0 then a else Plus (N i) a)" | "plus a (N i) = (if i=0 then a else Plus a (N i))" | "plus a\<^sub>1 a\<^sub>2 = Plus a\<^sub>1 a\<^sub>2" text_raw\<open>}%endsnip\<close>
lemma aval_plus[simp]: "aval (plus a1 a2) s = aval a1 s + aval a2 s" apply(induction a1 a2 rule: plus.induct) apply simp_all (* just for a change from auto *) done
text_raw\<open>\snip{AExpasimpdef}{2}{0}{%\<close> fun asimp :: "aexp \ aexp" where "asimp (N n) = N n" | "asimp (V x) = V x" | "asimp (Plus a\<^sub>1 a\<^sub>2) = plus (asimp a\<^sub>1) (asimp a\<^sub>2)" text_raw\<open>}%endsnip\<close>
text\<open>Note that in \<^const>\<open>asimp_const\<close> the optimized constructor was
inlined. Making it a separate function\<^const>\<open>plus\<close> improves modularity of
the code and the proofs.\<close>
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