https://github.com/sympy/sympy/issues/9135

On Tuesday, May 19, 2015 at 2:33:03 PM UTC-6, Jonathan Lindgren wrote:
>
> Well, that is exactly the problem, and what I think is a bug....it should 
> not work like that..
>
> Den tisdag 19 maj 2015 kl. 22:00:46 UTC+2 skrev Alexander Lindsay:
>>
>>  Not to hijack this post, but why does subs(phi(z,t).diff(z),...) work 
>> when phi(z,t).diff(t,2) does not contain any derivatives of phi with 
>> respect to z? It seems like this substitution is saying that 
>> phi(z,t).diff(z) = phi(z,t)
>>
>> I'm new to sympy, so I apologize if this is a stupid question.
>>
>> Alex
>>
>> On 05/19/2015 12:16 PM, Jonathan Lindgren wrote:
>>  
>> I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have 
>> some very strange behaviour with subs. The following code
>>
>> from sympy.abc import phi
>> import sympy as sp
>>
>> z=sp.Symbol('z')
>> t=sp.Symbol('t')
>>
>> sp.pprint((phi(z,t).diff(t,2)).subs(phi(z,t).diff(z),sp.Symbol('b')(z,t)).expand())
>>
>>
>> gives me the output     
>>   2         
>>  ∂          
>> ───(b(z, t))
>>   2         
>> ∂t   
>> but I would expect the output
>>   2         
>>  ∂          
>> ───(φ(z, t))
>>   2         
>> ∂t  
>>
>> This was working perfectly in my previous version of sympy.
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