Does the 1-loop correction for $g\bar\psi\psi$ term leads to the counterterms for additional terms in the Lagrangian?

Question

I have the Lagrangian in 4 dimensions: $$ L = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\sum_{i =1,2}\bar\psi_i(i\not\partial-m)\psi_i-g\phi\bar\psi_i\psi_i. $$

Assuming there are $n$ external scalar lines, and no external fermion lines, then the superficial degree of divergence could be found as $\omega = 4-n$.

Suppose $n= 3$, I can draw this divergent diagram, with the counterterms (a) and (b):

If I now add another term $g_3\phi^3$ to this Lagrangian, we can draw another divergent graph (c) with the same superficial degree of divergence. My question is does the counterterm stay the same if I add this term? I'm not quite sure how to make sense of this.

Actually, with $n = 2,3,4$, we will have this same question. Is it right if I say the 1-loop correction for the $g\bar\psi\psi$ term leads to the need of counterterms for $\phi^2$, $\phi^3$, and $\phi^4$?

Answer 1

Yes, one must for consistency as a minimum include all possible renormalizable terms that are not excluded by symmetry, cf. my related Phys.SE answer here. The $\phi^n$ vertex with $n=1,2,3,4,$ is e.g. generated from a 1-loop diagram of Yukawa $\phi\bar{\psi}\psi$ vertices with the fermion running in a loop.