 Yablo’s paradox relies on an infinite sequence of claims \(A_0\), \(A_1\), \(A_2\), …, where each \(A_i\) says that all of the ‘greater’ \(A_k\) (i.e., the \(A_k\) such that \(k \gt i)\) are untrue. (In other words, each claim says of the rest that they’re all untrue.) Since we have an infinite sequence, this version of the Liar paradox appears to avoid the sort of circularity apparent in the previous examples; however, contradiction still seems to emerge. If \(A_0\) is true, then all of the ‘greater’ \(A_k\) are untrue, and a fortiori \(A_1\) is untrue. But, then, there is at least one true \(A_k\) greater than \(A_1\) (i.e., some \(A_k\) such that \(k \gt 1)\), which contradicts \(A_0\). Conversely, if \(A_0\) is untrue, then there’s at least one true \(A_k\) greater than \(A_0\). Letting \(A_m\) be such a one (i.e., a truth greater than \(A_0)\), we have it that \(A_{m+1}\) is untrue, in which case there’s some truth greater than \(A_{m+1}\). But this contradicts \(A_m\). What we have, then, is that if \(A_0\) (the first claim in the infinite sequence) is true or untrue, then it is both. And this, as in the other cases, is a contradiction.