Properties of the Shifting Theorem

Question

The Shifting Theorem states that $$\text{If} \ \ H(t)=\begin{cases} 0 & t<0 \\ 1 & t>0 \\ \end{cases}, \ \ \text{then} \ \ \mathcal{L}(f(t-c)H(t-c))=e^{-cs}F(s).$$

Does this mean that $$\mathcal{L}((-1+t)H(t-1))\equiv -\mathcal{L}((t-1)H(t-1))?$$ I'm unsure if the Heaviside function, $H(t-1)$, is affected by this change in sign.

Answer 1

$\newcommand{\L}{\mathcal{L}}$Yes, the last equation is correct, since $\L(-f(t))=-\L(f(t))$, just using linearity of the Laplace transform.