Forced reflection: $c(m+1-d) = c(m+1+d)$ for every $d \in [1, m]$ with $c(d) = c(m+1)$ · Prove2Me

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Forced reflection: $c(m+1-d) = c(m+1+d)$ for every $d \in [1, m]$ with $c(d) = c(m+1)$ · Prove2Me