How can I calculate the equivalent resistance of this circuit with resistors in parallel with wires? [closed]

Problem:

Calculate the equivalent resistance, $$R_\text{eq} ,$$ of this circuit:

$$\hspace{50px}$$ .

My solution attempt

1. Тhe $$12 \, \Omega$$ resistor and the $$6\, \Omega$$ resistor are in parallel, so$$R = {\left(\frac{1}{12\,\Omega} + \frac{1}{6\,\Omega} \right)}^{-1} = 4\,\Omega \,, \tag{1}$$

reducing the circuit diagram to:
$$\hspace{50px}$$ .

2. The $$4 \,\Omega$$ resistor and the $$12 \,\Omega$$ resistor are in series, so$$R = 12\,\Omega + 4\,\Omega = 16\,\Omega \,, \tag{2}$$

reducing the circuit diagram to:
$$\hspace{50px}$$ .

I am not sure of the answer, and I am so confused right now.

I am totally new to circuits and any help will be much appreciated.

closed as off-topic by ZeroTheHero, John Rennie, stafusa, Jon Custer, ACuriousMind♦Aug 14 at 17:23

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• Is the top circuit diagram exactly as it appears in the book? – Bob D Aug 13 at 20:42
• @BobD Yep , it's exactly the same – AmirWG Aug 13 at 20:45
• The reason I ask is the 12 ohm resistor on the left is short-circuited by the wire and can be replaced by a wire. Then you have a parallel combination of 6 ohm and 12 ohms and that would give you an answer different than the book. – Bob D Aug 13 at 20:49
• I see 4 Ohms also – Dale Aug 13 at 20:50
• how do you tell if a resistor is short-circuited or not ? – AmirWG Aug 13 at 20:51 