Vitalik Buterin: @MTabarrok 归纳证明很容易改写为矛盾证明。 “P(0) 为真，P(x) 暗示 P(x+1)，因此 P(any +ve int) 为真” 变成 “假设 P(x) 并不总是正确的。让 z 是最小的 x，其中 P(z) 为假。但是 P(z-1) 为真，所以 P(z) 为真，矛盾！”

原推：@MTabarrok Proof by induction is easy to reword as proof by contradiction.

“P(0) is true, P(x) implies P(x+1), therefore P(any +ve int) is true”

Becomes

“suppose P(x) is not always true. Let z be smallest x where P(z) is false. But then P(z-1) is true, so P(z) is true, contradiction!”
https://twitter.com/VitalikButerin/status/1523506281549807616