Martingale and quadratic covariance

Question

Let B a standard Brownian motion. $$M^i_t=\int_0^t\phi^n_t \, dB^i_s = \sum_{i=1}^{K_n-1}U_{t_{i-1}^n}(B_{t_{i}^n}^i-B_{t_{i-1}^n}^i) $$ $$N^i_t=\int_0^t\psi^n_t \, dB^j_s = \sum_{i=1}^{K_n-1}V_{t_{i-1}^n}(B_{t_{i}^n}^j-B_{t_{i-1}^n}^j) $$ In a proof that i'm following i'm not able to understand why, if I prove that $M^i_tN^i_t$ is a martingale, then $[M^i,N^j]_t=0$ for all $t>0$ . Where $[M^i,N^j]$ is the quadratic covariance)