My textbook said that for a venn diagram code of length 7, which has weight 3 and minimum distance 3,
1) if the transmitted code word is x and the received string is y and under the assumption that at most one error occurs, the distance from y to any other codeword is at last 2 from the triangle inequality, and
2)if the assumption for the number of error holds, the venn diagram code can correct any error.
Please check if my reasoning for 1) and 2) are correct.
1)$d(x,z) \le d(x,y)+d(y,z)\\\Rightarrow d(y,z) \ge d(x,z)-d(x,y)\ge3-1=2$, since $d(x,z) \ge 3$ and $d(x,y)=1$.
2) The distance from y to any other code is at least two. This means if y was the result of error caused in another message, there would have to be at least two errors, contradicting the assumption that there is at most one error.
This is also known as Hamming $(7,4)$ code.
Yes, 1) follows from the reverse triangle inequality: the distance from received signal to any codeword that was not sent is at least the code's minimal distance minus the number of errors.
2) is correct: if there is at most one error, the decoder will not pick a wrong codeword, by virtue of 1).