Re: [isabelle] Dealing with lets in Isar

```Hi Christian,

```
I don't completely understand why this is related to let. A let just introduces an abbreviation. However, in your example, you are more concerned with patterns such as tuples, but
```
let (x, y) = e in f x y

is syntactic sugar for

Let e (prod_case (%x y. f x y))

```
So, do you also want to introduce variables if there's no pattern-matching involved, such that "P (let x = e in f x)" gets transformed into "!!x. x = e ==> P (f x)"?
```
In principle, you can use Isabelle's splitter:

lemma Let_prod_split_asm:
"P (Let e (prod_case f)) = (~ (EX x y. (x, y) = e & ~ P (f x y)))"

lemma Let_split_asm: "P (Let e f) = (~ (EX x. x = e & ~ P (f x)))"

lemma
assumes "P (let (x, y) = e1; (u, v) = e2; (w, k) = e3 in f x y u v w k)"
shows "thesis"
using assms
apply -
proof(split Let_prod_split_asm Let_split_asm)+
case (goal1 x y u v w k)

With some ML programming, it should be easy to make this more readable.

Hope this helps,
Andreas

On 29/11/13 16:55, Christian Urban wrote:
```
```
Dear All,

We have a question about how to deal conveniently with
Lets in Isar proofs. Some of our definitions contain many
nested lets, as in

definition
"foo f e1 e2 e3 == let (x, y) = e1; (u, v) = e2; (w, k) = e3 in f x y u v w k"

The problem is that we have foo unfolded in our premises, say

lemma
assumes "P (let (x, y) = e1; (u, v) = e2; (w, k) = e3 in f x y u v w k)"
shows "thesis"

We could prove such lemmas with

lemma
assumes "P (let (x, y) = e1; (u, v) = e2; (w, k) = e3 in f x y u v w k)"
shows "thesis"
proof -
from assms
obtain x y u v w k
where "P (f x y u v w k)" and
"(x, y) = e1" and "(u, v) = e2" and "(w, k) = e3" by ...

but we are wondering whether there is a simpler way. Essentially,
we want to avoid having to fix the variables and to avoid to write
the equations explicitly. We rather prefer the way how the cases-construction
works, which automatically fixes variables and puts such equations
into the context, for example

lemma
assumes "P (n:: nat)"
shows "thesis"
proof (cases n)

In this example you have "n = 0" in the "first" case and fix nat/n = Suc nat
in the "second" case. Is there something similar for many nested lets?

If not, how would we have to go about implementing (presumably in ML)
such a construction?

Best wishes,
Christian

```
```

```

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