Lemma: For the two-step Adams-Bashforth (AB) method $$T(t)=\frac{5}{12} y'''(t)h^2+\mathcal{O}(h^3),$$ where $T(t)$ denotes the local truncation error.
By definition, $$T(t)=\frac{y(t+h)-y(t)}{h}-f(t,y(t)).$$ For the AB method, $$T(t)=\frac{y(t+h)-y(t)}{h}-\sum_{j=0}^{r}b_jf(t-jh,y(t-jh)).$$ The lemma is the result for $r=1$. By Taylor's theorem, \begin{align} \frac{y(t+h)-y(t)}{h}&=\frac{\left(y(t)+y'(t)h+y''(t)h^2/2+y'''(t)h^3/6+\mathcal{O}(h^4)\right)-y(t)}{h} \tag{1}\\ &=y'(t)+\frac{y''(t)}{2}h+\frac{y'''(t)}{6}h^2+\mathcal{O}(h^3) \tag{*} \end{align}For $r=1$: \begin{align} \sum_{j=0}^{1}b_jf(t-jh,y(t-jh))&=b_0f(t,y(t))+b_1f(t-h,y(t-h)) \\ &=\frac{3}{2}f(t,y(t))-\frac{1}{2}f(t-h,y(t-h)) \ \ \ (b_0=3/2, \ b_1=-1/2)\\ &=\frac{3}{2}y'(t)-\frac{1}{2}y'(t-h) \ \ \ (y'(t)=f(t,y(t)) ) \\ &=\frac{3}{2}y'(t)-\frac{1}{2}\left(y'(t)-y''(t)h+\frac{1}{2}y'''(t)h^2+\mathcal{O}(h^3)\right) \tag{2} \\ &=y'(t)+\frac{y''(t)}{2}h-\frac{y'''(t)}{4}h^2+\mathcal{O}(h^3) \tag{**} \end{align} Now the result follows by subtracting $(**)$ from $(*)$, but I don't understand how Taylor's theorem was used in steps $(1)$ and $(2)$, i.e. the expansions of $y(t+h)$ and $y'(t-h)$.