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From the given circuit diagram:

1. The node between $R_1$ and $R_3$ is directly connected to the top node above $R_6$ via an ideal conducting wire. Let this be node C.
2. The node between $R_2$, $R_3$, $R_4$, and $R_5$ is a common node. Let this be node D.
3. Resistors $R_5 = 10\ \Omega$ and $R_6 = 10\ \Omega$ are connected in series across nodes C and D.
   $R_{56} = R_5 + R_6 = 10 + 10 = 20\ \Omega$
4. Across nodes C and D, the resistors $R_3 = 10\ \Omega$, $R_4 = 10\ \Omega$, and $R_{56} = 20\ \Omega$ are in parallel:
   $\frac{1}{R_{CD}} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_{56}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4}\ \Omega^{-1}$
   $R_{CD} = 4\ \Omega$
5. The total equivalent resistance between terminals A and B is the series combination of $R_1$, $R_{CD}$, and $R_2$:
   $R_{AB} = R_1 + R_{CD} + R_2 = 5\ \Omega + 4\ \Omega + 4\ \Omega = 13\ \Omega$

Answer: $13\ \Omega$