Let $(R, \mathfrak m)$ be a Noetherian local ring of depth $t\ge 1$. So for any finitely generated free $R$-module $F$, we have $\operatorname {depth}(F)=\operatorname {depth}(R)=t$.
My question:
If $0\ne M$ is a finitely generated $R$-module which fits into an exact sequence $0\to M \to F_{i-1}\to \dots\to F_1\to F_0$, where $1\le i\le t$ and $F_i$ are finitely generated free $R$-modules, then how to show that $\operatorname {depth}(M)\ge i$ ?
I can easily show this if $t=1$ or $2$.
Also, for $i=1$ (irrespective of $t$) , since $0\to M\to F_0$ and $\operatorname {depth}(F_0)=\operatorname {depth}(R)=t\ge 1$ and submodules of modules of depth $\ge 1$ always has depth $\ge 1$, so in that case $\operatorname {depth}(M)\ge 1$.
But I'm unable to do anything for higher values of $t$.
Please help.