Basic Ideal Op Amp Circuit Analysis
Let's perform the calculations to verify the explanation.

Since \$v1=v2\$ the 30k&Omega; resistor has the entire 6V from the generator across it, hence the current:

$$
i=\frac{6V}{30k\Omega}=0.2mA
$$

Since the opamp inputs (ideally) draw no current, this is the same current flowing in the 10k&Omega; resistor. Therefore, because the current direction is from ground to \$v1\$, we have: 

$$
v_1 = - i \cdot 10k\Omega = -0.2mA \cdot 10k\Omega = -2V
$$

We already know that \$v1=v2=-2V\$ because of the opamp action (virtual short between inputs) due to negative feedback. Moreover the same current \$i\$ flows in the 20k&Omega; resistor, hence we can write:

$$
v_3 = v_2 - 20k\Omega \cdot i 
= -2V - 20k\Omega \cdot 0.2mA 
= -2V - 4V = -6V
$$

In the end your conclusion is right. I hope this explicit calculation procedure dissipates any residual doubt you have.