The temporal difference (TD) error from eligibility trace update rule is,

\[\Delta V_t^{TD}(s) = \alpha \delta_t e_t(s)\]

where, \(\delta_t = r_{t+1} + \gamma V(s_{t+1}) - V(s_t)\) and \(e_t(s) = \gamma \lambda e_{t-1}(s) + I_{ss_t}\) with \(I_{ss_t} = 1\) if \(s=s_t\) and \(0\) otherwise.

Expanding \(e_t(s)\) we obtain

\(e_t(s) = I_{ss_t} + \gamma \lambda e_{t-1}(s)\) \(e_t(s) = I_{ss_t} + \gamma \lambda \left(\gamma \lambda e_{t-2}(s) + I_{ss_{t-1}}\right)\) \(e_t(s) = I_{ss_t} + \gamma \lambda I_{ss_{t-1}} + \gamma^2 \lambda^2 I_{ss_{t-2}} + \gamma^3 \lambda^3 I_{ss_{t-3}} + \dots\) \(e_t(s) = \sum_{k=0}^t (\gamma \lambda)^{t-k} I_{ss_k}\)

The sum of TD errors over the trajectory in backward view for a state $s$ is,

\(\sum_{t=0}^{T-1} \Delta V_t^{TD}(s) = \sum_{t=0}^{T-1} \alpha \delta_t e_t(s)\) \(\sum_{t=0}^{T-1} \Delta V_t^{TD}(s) = \sum_{t=0}^{T-1} \alpha \delta_t \left( \sum_{k=0}^{t} (\gamma \lambda)^{t-k} I_{ss_k}\right)\)

Interchanging \(t\) with \(k\) and \(k\) with \(t\) on the RHS, we obtain

\[\sum_{t=0}^{T-1} \Delta V_t^{TD}(s) = \sum_{k=0}^{T-1} \alpha \delta_t \left( \sum_{t=0}^{k} (\gamma \lambda)^{k-t} I_{ss_k}\right)\]

Expanding the RHS results in,

\(\sum_{t=0}^{T-1} \Delta V_t^{TD}(s) = \alpha \delta_0 (\gamma \lambda)^0 I_{ss_0} + \delta_1 ( \gamma \lambda I_{ss_0} + (\gamma \lambda)^0 I_{ss_1}) +\) \(+ \delta_2 ( \gamma^2 \lambda^2 I_{ss_0} + \gamma \lambda I_{ss_1} + (\gamma \lambda)^0 I_{ss_2})\)

\[+ \delta_3 ( \gamma^3 \lambda^3 I_{ss_0} + \gamma^2 \lambda^2 I_{ss_1} + \gamma \lambda I_{ss_2} + (\gamma \lambda)^0 I_{ss_3}) + \dots\] \[= \alpha \left( I_{ss_0} \sum_{k=0}^{T-1} (\gamma \lambda)^k \delta_k + I_{ss_1} \sum_{k=1}^{T-1} (\gamma \lambda)^{k-1} \delta_k + \dots \right)\] \[= \alpha \sum_{t=0}^{T-1} I_{ss_t} \left( \sum_{k=t}^{T-1} (\gamma \lambda)^{k-t} \delta_k \right)\]

Now, from the foward view the update rule is,

\[\Delta V_t^{\lambda}(s_t) = \alpha (L_t^{\lambda} - V_t(s_t))\]

where, \(L_t^{\lambda}\) is the weighted sum of n-step returns

Dividing the above equation by $\alpha$, we get

\(\frac{1}{\alpha} \Delta V_t^{\lambda}(s_t) = L_t^{\lambda} - V_t(s_t)\) \(= - V_t(s_t) + (1-\lambda) (R_{t+1} + \gamma V_t(s_{t+1})) + (1-\lambda)\)