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 
>
>

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