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This is from a past paper exam I am revising. Please can someone explain if this is antisymmetric or not. According to the answer it says it isn't, but I can't for the life of me understand why.

Consider the set N + of all non-empty finite sequences of natural numbers – for example, [1, 1], [1, 2, 1] and [2, 0, 2, 0, 0] are all elements of N +. Let R be the binary relation over N + defined by (s, t) ∈ R if and only if s1 ≤ t1 where s1 and t1 are the first numbers of the sequences s and t, respectively. For example, ([1, 1], [1, 2, 1]) ∈ R but ([2, 0, 2, 0, 0], [1, 1]) ∈/ R.

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It is not antisymmetric because, for instance, $[1\ \ 1]\mathrel R[1\ \ 2]$ and $[1\ \ 2]\mathrel R[1\ \ 1]$ , but you don't have $[1\ \ 1]=[1\ \ 2]$ .

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  • $\begingroup$ Can you explain how [1 2] R [1 1] is a relation? I thought R was the less than or equal sign? If it is symmetric, how are they now equal? $\endgroup$ – Jake Jackson May 2 at 21:50
  • $\begingroup$ We have $[1\ \ 2]\mathrel R[1\ \ 1]$ because the first element of $[1\ \ 2]$ (which is $1$) is smaller than or equal to the first element of $[1\ \ 1]$ (which is also $1$). I don't understand your other questions. $\endgroup$ – José Carlos Santos May 2 at 21:54

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