Can you predict cancellations by supersymmetry?

Question

I'm studying supersymmetry breaking using Turing's textbook, Modern Supersymmetry (pg 91). He begins with the O'Raifeartaigh model,
\begin{equation} 
W = g \Phi _1 \left( \Phi _3 ^2 - m ^2  \right) + M \Phi _2 \Phi _3 
\end{equation} 
which gives the explicit Lagrangian,
\begin{align} 
\Delta {\cal L} & = \sum _i ( i \bar{\psi}  _i \bar{\sigma}^{\mu}\partial_\mu \psi _i+ \partial _\mu \phi _i ^\dagger \partial ^\mu \phi _i )  -  M ( \psi _2 \psi _3 + h.c. ) \\  & - g ( \phi _1  \psi _3 \psi _3 + h.c. ) - 2g ( \phi _3 \psi _1 \psi _3 + h.c. )     \\  &  - M ^2 \phi _3 ^\ast \phi _3  - M ^2 \phi _2 ^\ast \phi _2  +g ^2 m ^2 (  \phi _3 ^2 + h.c. )  - 2 g  M ( \phi _1 \phi _2 \phi _3 + h.c. ) \\ & - 4 g ^2 \phi _1 \phi _1 ^\ast  \phi _3 \phi _3 ^\ast  \end{align} 

where $\psi _1$ is the Goldstino. It's superpartner $\phi _1$ is also massless at tree level. Turing then goes on to look for its one loop mass correction which involves calculating the diagrams shown here (I can't figure out how to embed the image directly, if you know how feel free to edit the question).

He then says that ``The correction to the $\phi _1$ mass from the top three graphs... vanish by SUSY.'' Is this statement obviously true or just an interesting observation one could make after the fact? In other words, could I have made this claim without calculating it explicitly?

Comments

This is a general answer and not specific to the mentioned example. If one works with superfields and supergraphs, then bose-fermi cancellations are easy to see as several Feynman diagrams combine to form a single supergraph. Supergraphs are discussed in Wess & Bagger.

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Thanks, I'll keep that in mind. I have yet to find the time to learn the supergraph formalism.