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The transformation of the holomorphic strong-coupling scale is fixed by the mixed $SU(2{)}^{2}U(1)$ anomaly. The baryon-number contributions of $\Phi$ and $\tilde{\Phi }$ cancel, so ${\Lambda }^{b_0}$ is neutral under $U(1{)}_B$. Under $U(1{)}_A$, all $2N_f$ doublets have charge one and Dynkin index one, giving axial charge $2N_f$. For the R-symmetry, each matter fermion has charge $(1-2/N_f)-1=-2/N_f$, so the matter contribution is $-4$; the gaugino has R-charge one and contributes $I(\mathrm{adj})=4$, giving zero total anomaly. Thus

The mixed $SU(N{)}^{2}U(1{)}_R$ anomaly receives $I(\mathrm{adj})=2N$ from the gaugino. A chiral multiplet of R-charge $r$ contains a fermion of charge $r-1$. Withthe symmetric-tensor fermion contributeswhile every antifundamental fermion contributes zero. The total is $2N-2N=0$, so this is a nonanomalous R-symmetry of the quantum theory.

In the proposed Spin(8) theory, the gaugino contributes $I(\mathrm{adj})=6$ to the mixed gauge-R anomaly. The $p$ fields $q$ have superfield R-charge one, so their fermions contribute zero. The spinor $t$ has fermion R-charge $-5-1=-6$ and index one, giving $-6$. The singlets $M$ and $U$ do not enter the gauge anomaly. Henceand the proposed R-symmetry survives quantization.