how to make an example to test the rev_app immediately after lemma proved. an starting example for custom lemma - isabelle

Question

expect to use the subgoal to run the list which defined by let? aa = [1,2]
and run rev_app on this aa and show the value as [2,1]
theory Scratch2
imports Datatype
begin
datatype 'a list = Nil ("[]")
| Cons 'a "'a list" (infixr "#" 65)
(* This is the append function: *)
primrec app :: "'a list => 'a list => 'a list" (infixr "#" 65)
where
"[] # ys = ys" |
"(x # xs) # ys = x # (xs # ys)"
primrec rev :: "'a list => 'a list" where
"rev [] = []" |
"rev (x # xs) = (rev xs) # (x # [])"
primrec itrev :: "'a list => 'a list => 'a list" where
"itrev [] ys = ys" |
"itrev (x#xs) ys = itrev xs (x#ys)"
value "rev (True # False # [])"
lemma app_Nil2 [simp]: "xs # [] = xs"
apply(induct_tac xs)
apply(auto)
done
lemma app_assoc [simp]: "(xs # ys) # zs = xs # (ys # zs)"
apply(induct_tac xs)
apply(auto)
done
(1 st trial)
lemma rev_app [simp]: "rev(xs # ys) = (rev ys) # (rev xs)"
apply(induct_tac xs)
thus ?aa by rev_app
show "rev_app [1; 2]"
(2nd trial)
value "rev_app [1,2]"
(3 rd trial)
fun ff :: "'a list ⇒ 'a list"
where "rev(xs # ys) = (rev ys) # (rev xs)"
value "ff [1,2]"
thus ?aa by rev_app
show "rev_app [1; 2]"
end

Answer 1

Firstly, you need the syntax for list enumeration (I just picked it up in the src/HOL/List.thy file):
syntax
-- {* list Enumeration *}
"_list" :: "args => 'a list" ("[(_)]")
translations
"[x, xs]" == "x#[xs]"
"[x]" == "x#[]"
Then, is one of the following what you're searching for ?
Proposition 1:
lemma example1: "rev [a, b] = [b, a]"
by simp
This lemma is proved by applying the definition rules of rev that are used by the method simp to rewrite the left-hand term and prove that the two sides of the equality are equal. This is the solution I prefer because you can see the example is satisfied even without evaluating it with Isabelle.
Proposition 2:
value "rev [a, b]" (* return "[b, a]" *)
Here and in Proposition 3, we just uses the command value to evaluate rev.
Proposition 3:
value "rev [a, b] = [b, a]" (* returns "True" *)
This lemma is not used by the previous propositions:
lemma rev_app [simp]: "rev(xs # ys) = (rev ys) # (rev xs)"
apply (induct_tac xs)
by simp_all
Notes:
As a general principle, you shouldn't import the "Datatype" package alone, but import "Main" instead.
In your 1st attempt you're mixing the "apply" (apply ...) and the "structured proof" (thus ...) styles
"thus ?aa" makes no sense if "?aa" is "[1,2]" as the argument of "thus" should be a subgoal, ie. a proposition with a boolean value.
To evaluate, the command "value" uses ML execution or if this fails, normalisation by evaluation.
In example1, you can use a custom proof and thus lemmas (for example: by (simp add:rev_app)