# Is minimal cover under symmetric 3-deduction NP-complete?

Forgive me if this problem is known by another name, I do not know any references for it.

Symmetric deduction. An equation $$e \in E$$ is a subset of variables $$V$$ such that knowing $$|e| - 1$$ of the variables allows you to deduce the missing one.

Minimal cover under symmetric deduction. Does there exist a set of $$k$$ variables such that their closure under symmetric deduction with equations $$E$$ equals all variables $$V$$?

This problem is proven NP-complete in this question. There is a limited version of this problem:

Minimal cover under symmetric $$d$$-deduction. Same as before, except $$\forall e\in E: |e| \leq d$$. That is, each equation has degree $$d$$ or lower.

This problem is trivial for $$d = 2$$ (choose one variable per connected component), and NP-hard for $$d = 4$$ as shown by Thinh D. Nugyen here.

What about $$d = 3$$?