How to multiply out a statement form?

Question

I got this form:

(not M or V) and (A or not M) and (not B or M) and (B or V) and (A or not V) and (not A or B)

Or: $$(\neg M\vee V) \wedge (A\vee\neg M) \wedge (\neg B \vee M) \wedge (B\vee V) \wedge (A\vee\neg V)\wedge(\neg A \vee B)$$

Can anyone explain me how to solve it step by step? The result should be this:

A and B and M and V: $$A\wedge B \wedge M \wedge V$$

Wolfram Alpha link.

I dont know how to simplify the statement

Answer 1

$$(\neg M\vee V) \wedge \color{blue}{(A\vee\neg M)} \wedge (\neg B \vee M) \wedge \color{red}{(B\vee V) \wedge} \color{blue}{(A\vee\neg V)}\wedge \color{red}{(\neg A \vee B)}\tag{1}$$

$$\color{blue}{[A \lor (\lnot V \land \lnot M)]} \land \color{red}{[B \lor (\lnot A \land V)]} \land \color{green}{(\lnot B \lor M) \land (\lnot M \lor V)}\tag{2}$$

$${[A \lor (\lnot V \land \lnot M)]} \land {[B \lor (\lnot A \land V)]} \land \color{green}{(B \rightarrow M) \land ( M \rightarrow V)}\tag{3}$$

$$\vdots$$

$$A \land B\land M\land V\tag{result}$$

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?

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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:
$$(\neg M\vee V) \wedge {(A\vee\neg M)} \wedge (\neg B \vee M) \wedge {(B\vee V) \wedge} {(A\vee\neg V)}\wedge {(\neg A \vee B)}$$ $$\iff A \land B \land M \land V$$

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