Answer by saolof for In choiceless constructivism: If $f'=0$ then is $f$...
Here's a stab at the question that attempts to make use of the locale of real numbers. (If someone has published something similar, please point me towards it). The reasoning behind this is that if you...
View ArticleAnswer by Andrej Bauer for In choiceless constructivism: If $f'=0$ then is...
For the record, I provide here a proof of the constancy principle from the principle of open induction. Recall:Principle of open induction:Let $U \subseteq [0,1]$ be an open set such that$$\forall x...
View ArticleAnswer by François G. Dorais for In choiceless constructivism: If $f'=0$ then...
This answer is incorrect since the function $f\colon\mathbb R \to\mathbb R$ is not computable. That said, it is possible that a similar idea could provide a counterexample.There is a computable closed...
View ArticleAnswer by Franka Waaldijk for In choiceless constructivism: If $f'=0$ then is...
For a real function $f$ with continuous derivative $f'$ we have the following identity which should not require any choice to prove:$$ f(x) = f(0) + \int_0^x f'(y)dy \ \ (\mbox{all}\ x\in\mathbb{R})...
View ArticleIn choiceless constructivism: If $f'=0$ then is $f$ constant?
Prove, without any Choice principles or Excluded Middle, that if a pointwise differentiable function has derivative $0$ everywhere, then it is constant. The function in this case maps $\mathbb R$ to...
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