The proof proceeds by induction. For all $n \in \Z_{\le 0}$, let $P \left({n}\right)$ be the proposition: :$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ This can equivalently be written: For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$F_{m - n} = F_{m - 1} F_{-n} + F_m F_{-n + 1}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{m - 1} F_0 + F_m F_1       | r = F_m       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{eqn | r = F_{m + 0}       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{m - 1} F_{-1} + F_m F_0       | r = F_{m - 1} F_{-1}       | c = {{Defof|Fibonacci Number}}: $F_0 = 0$ }} {{eqn | r = F_{m - 1} \left({-1}\right)^0 F_1       | c = Fibonacci Number with Negative Index }} {{eqn | r = F_{m - 1} F_1       | c =  }} {{eqn | r = F_{m - 1}       | c = {{Defof|Fibonacci Number}}: $F_1 = 1$ }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ and $P \left({k - 1}\right)$ are true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$F_{m - k} = F_{m - 1} F_{-k} + F_m F_{-k + 1}$ and: :$F_{m - \left({k - 1}\right)} = F_{m - 1} F_{-\left({k - 1}\right)} + F_m F_{-\left({k - 1}\right) + 1}$ from which it is to be shown that: {{begin-eqn}} {{eqn | l = F_{m - \left({k + 1}\right)}       | r = F_{m - 1} F_{-\left({k + 1}\right)} + F_m F_{-\left({k + 1}\right) + 1}       | c =  }} {{eqn | ll= \leadsto       | l = F_{m - k - 1}       | r = F_{m - 1} F_{-k - 1} + F_m F_{-k}       | c =  }} {{end-eqn}} === Induction Step === First we note that: {{begin-eqn}} {{eqn | l = F_{m - k}       | r = F_{m - 1} F_{-k} + F_m F_{-k + 1}       | c =  }} {{eqn | ll= \iff       | l = F_{m - k}       | r = F_{m - 1} F_{-k} + F_m F_{-\left({k - 1}\right)}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = F_{m - \left({k - 1}\right)}       | r = F_{m - 1} F_{-\left({k - 1}\right)} + F_m F_{-\left({k - 1}\right) + 1}       | c =  }} {{eqn | ll= \iff       | l = F_{m - k + 1}       | r = F_{m - 1} F_{-k + 1} + F_m F_{-k + 2}       | c =  }} {{end-eqn}} This is the induction step: {{begin-eqn}} {{eqn | l = F_{m - k - 1}       | r = F_{m - k + 1} - F_{m - k}       | c = {{Defof|Fibonacci Number for Negative Index}} }} {{eqn | r = F_{m - \left({k - 1}\right)} - F_{m - k}       | c =  }} {{eqn | r = \left({F_{m - 1} F_{-\left({k - 1}\right)} + F_m F_{-\left({k - 1}\right) + 1} }\right) - \left({F_{m - 1} F_{-k} + F_m F_{-k + 1} }\right)       | c = Induction Hypothesis }} {{eqn | r = F_{m - 1} \left({F_{-\left({k - 1}\right)} - F_{-k} }\right) + F_m \left({F_{-\left({k - 1}\right) + 1} - F_{-k + 1} }\right)       | c =  }} {{eqn | r = F_{m - 1} \left({F_{-k + 1} - F_{-k} }\right) + F_m \left({F_{-k + 2} - F_{-k + 1} }\right)       | c =  }} {{eqn | r = F_{m - 1} F_{-k - 1} + F_m F_{-k}       | c = {{Defof|Fibonacci Number for Negative Index}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\le 0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ </onlyinclude>