
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Given a function $y=f(x)$, where $x \in [0,+\infty)$, for a given non-zero real number $a$, there always exists a non-zero constant $T$ such that for any real number $x$ in the domain $[0,+\infty)$, $af(x)=f(x+T)$ always holds. In this case, $T$ is called the quasi-period of $f(x)$, and the function $y=f(x)$ is an $a$-level quasi-periodic function on $[0,+\infty)$. If the function $y=f(x)$ is a 2-level quasi-periodic function defined on the interval $[0,+\infty)$ with $T=2$, and when $x \in [0,2)$, $f(x)= \begin{cases} \frac {1}{2}-2x^{2}, & 0\leqslant x\leqslant 1 \\ f(2-x), & 1 < x < 2 \end{cases}$. The function $g(x)=-2\ln x+ \frac {1}{2}x^{2}+x+m$. Determine the range of values for the real number $m$ such that there exists $x_{1} \in [6,8]$ and $x_{2} \in (0,+\infty)$ such that $g(x_{2})-f(x_{1})\leqslant 0$ holds.

Present the answer in LaTex format: \boxed{Your answer}
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