((n/n) == (n-1)/(n-1)) && Implies[((a/a) = (b/b)), (a = b)] && Implies[((n/n) == ((n-1)/(n-1))) && Implies[(a/a) == (b/b), (a == b)], (n == n - 1)] && Implies[n == n - 1, 1 == 2]

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