$(\neg M\vee V) \wedge \color{blue}{(A\vee\neg M)} \wedge (\neg B \vee M) \wedge \color{red}{(B\vee V) \wedge} \color{blue}{(A\vee\neg V)}\wedge \color{red}{(\neg A \vee B)}\tag{1}$
$\color{blue}{[A \lor (\lnot V \land \lnot M)]} \land \color{red}{[B \lor (\lnot A \land V)]} \land \color{green}{(\lnot B \lor M) \land (\lnot M \lor V)}\tag{2}$
${[A \lor (\lnot V \land \lnot M)]} \land {[B \lor (\lnot A \land V)]} \land \color{green}{(B \rightarrow M) \land ( M \rightarrow V)}\tag{3}$
$\vdots$
$A \land B\land M\land V\tag{result}$
Note: To "deduce" this, I've highlighted some initial steps:
- $(1) \to (2)$ using
- commutativity: $(P \lor Q) \equiv (Q \lor P)$ and $P \land Q \equiv Q \land P$,
- associativity: $P\land (Q \land R) \equiv (P\land Q)\land R$,
- distributive law (applied twice): $(P\lor Q) \land (P\lor R) \equiv P\lor (Q \land R)$, and
- $(2)\to (3)$ using the identity ($\lnot P \lor Q) \equiv (P\rightarrow Q$),
- It follows from $(3)$ (with more work needed to establish) that we must have $A$ and $B$; and since $B$, then also $M$; and since $M$, then also $V$. Why?
It also helps to use a truth-table, from which we can derive the conjunctive-normal form $(\text{result})$ of your expression given in $(1)$:

Note from the truth-table that the given expression evaluates to $\;T=$ true$\;$ only in the first row, if and only if $\;A,\text{ AND}\; B, \text{ AND}\;M, \text{ AND}\; V\;$ are all $\;T=$ true$\;$.
That is, we can conclude:
$(\neg M\vee V) \wedge {(A\vee\neg M)} \wedge (\neg B \vee M) \wedge {(B\vee V) \wedge} {(A\vee\neg V)}\wedge {(\neg A \vee B)}$ $\iff A \land B \land M \land V$