# Re: [isabelle] A bug in chained proof mode

```Dear Eddy,

this actually is the intended behavior, the reason of which is explained in:

G. Bauer and M. Wenzel. Calculational reasoning revisited - an
Isabelle/Isar experience. In R. J. Boulton and P. B. Jackson, editors,
Theorem Proving in Higher Order Logics: TPHOLs 2001, volume 2152 of
Lecture Notes in Computer Science. Springer-Verlag, 2001.

```
According to the isar-ref manual (see Section 2.2.4 "Calculational reasoning") your example would translate into
```
assume "e1 = e2"
have "f e1 = f e2" ...
note calculation = this (* "f e1 = fe2" *)
have "f e2 = f e2" ...
note calculation = <TRANS> [OF calculation this] (* "f e1 = f e2" *)
from calculation
have "f e1 = f e2"

```
where the crucial spot is <TRANS>. That is, the actual content of "calculation" depends on the chosen rule. And as explained in the above paper this is done by trying all declared (via [trans] and [sym]; those can be consulted via "print_trans_rules") lemmas while discarding results that are mere projections.
```
cheers

chris

```
Btw: *wispered* you should be careful when throwing around words like "bug" ;)
```
On 06/11/2014 07:03 AM, Eddy Westbrook wrote:
```
```Hi,

I think I have found a bug in the chained proof mode, that occurs for equality proofs if one of the proof used in the chain is a reflexive proof. As an example, the following proof fails, even though (as far as I understand the chain mode) it should succeed:

lemma chain_bug: "e1 = e2 ⟹ f e1 = f e2"
proof -
assume eq1: "e1 = e2"
have "f e1 = f e2" by (rule arg_cong[OF eq1])
also
have "f e2 = f e2" by (simp)
finally
have "f e1 = f e2".
qed

Although it may seem silly to write a proof that “f e2 = f e2”, the reason it comes up is that the proof is machine-generated, and sometimes it is easier to generate a proof that something equals itself rather than checking explicitly for that fact.

Anyway, just thought I would report this.

Thanks,
-Eddy

```
```

```

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