slowly with $k$ and ${p_\text{typ}}$ than with $\Lambda$. The results of such a fit may certainly be inconclusive; see extended remarks in Sect. \[sec:notesofnote\]. But if higher orders are *not* parametrically suppressed and the slope comes out *smaller* than the PC prediction $n+1$, then the EFT is necessarily not properly renormalised. As will be discussed in Sect. \[sec:principle\], a slope $\ge n+1$ does not suffice to demonstrate consistency or establish failure.

Equation is formulated in terms of renormalised quantities only and therefore holds for any regulator, but it is most useful for cutoffs: Answers in nonperturbative EFTs are usually found only numerically and for a *cutoff regulator*, i.e. for a regulator which explicitly suppresses high momenta $q\gtrsim\Lambda$ in loops. It is this case which we use to our advantage from now on.

Cutoffs are of course only sensible if all loop momenta are sampled which lie in the domain of validity of the EFT, i.e. if $\Lambda\gtrsim{\overline{\Lambda}_\text{EFT}}$. Only then can the coefficients ${\mathcal{C}}_n$ be expected to be of natural size relative to ${\overline{\Lambda}_\text{EFT}}$ (with the caveats mentioned around eq. ). Except for this, no particular assumption is necessary as to the size of $\Lambda$ relative to ${\overline{\Lambda}_\text{EFT}}$ in eq. . 