I afraid I can't help there. I am good at finding things with google.
This might help, maybe?
http://pyeda.readthedocs.io/en/latest/reference/boolalg/boolfunc.html
On 09/20/2017 12:57 PM, Chris Smith wrote:
I will need some help interpreting that. Here's my attempt:
|
>>>f =eq =ITE(x<1,x,Eq(y,1)).to_nnf();eq
(x∨x≥1)∧(y=1∨x<1)(x∨x≥1)∧(y=1∨x<1)
>>>eq.subs(x,0)
False
>>>eq.subs(x,1)
y =1
|
I'm not sure how I am supposed to add these mod 2. Maybe `(0 +
ITE(Eq(y,1), 1, 0)) % 2`? So does x change the value of f? The answer
depends on y, doesn't it? If y is 0 then the value of x doesn't change
f but it it is 1 then it does. So is the derivative wrt x `Eq(y, 1)`?
On Tuesday, September 19, 2017 at 4:06:04 PM UTC-5, brombo wrote:
On 09/19/2017 04:24 PM, Aaron Meurer wrote:
> I'm not sure the derivative really makes sense. I would either
error
> or leave it unevaluated.
>
> Aaron Meurer
>
> On Tue, Sep 19, 2017 at 9:19 AM, Chris Smith <[email protected]
<javascript:>> wrote:
>> In PR #13204 I encountered the question of what to do with the
derivative if
>> a Boolean. A Boolean is true or false but may contain
expressions within
>> relationals, For example, `x**2 < y` depends on x and y but the
Boolean
>> value is a "square wave" for this relational. Does a derivative
wrt x or y
>> even make sense? Should an error, 0 or something else be done
for a return
>> value?
>>
>> /c
>>
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See link -
https://www.encyclopediaofmath.org/index.php/Boolean_differential_calculus
<https://www.encyclopediaofmath.org/index.php/Boolean_differential_calculus>
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