# Re: [isabelle] Automatic simplification of lambda terms of tuples

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On 26/06/2015 16:43, Iulia Dragomir wrote:
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```Dear all,

I have the following lemma within a theory:

theory Test imports Main
begin

lemma lambda_true_top: "f (Î x . True) = F top"
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```
I assume you mean the same f/F on both sides. This does not cover abstraction over tuples, which is syntactic sugar for the combinator split :: "('a â 'b â 'c) â 'a Ã 'b â 'c". You also need
```
lemma lambda_pair_true_top: "f (Î (x,y) . True) = f top"

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However, the simplifier will not be able to use such rules, unless you instantiate the f with a concrete term.
```
Tobias

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```end

which I would like to use for lambda terms having different variables, like:
lemma "F (Î (x,y) . True) = F top" or
lemma "F (Î (x,y, (z, t), a, (b, c, d)) . True) = F top"
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```
```
```Now, if I use it as substitution rule, the goal is identical to the
lemmas to be proved and if I try to add it as simplification rule, a
failed application of proof method is raised.
How can I use lambda_true_top in this proof? Or how can I write my
original lemma such that it can be used to simplify a term like F (Î
(x,y, (z, t), a, (b, c, d)) . True) to F top in general? In my case F is
always a constant function.

Thanks.

Best regards,
Iulia Dragomir

```
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