(A.1): $\bar\omega_i^{II}(\beta^*/\beta_i) + A_i = \bar\omega_0(\beta^*/\beta_i)$ · Prove2Me

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Initializing...

(A.1): $\bar\omega_i^{II}(\beta^*/\beta_i) + A_i = \bar\omega_0(\beta^*/\beta_i)$ · Prove2Me