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Yablo's Paradox1

Selmer Bringsjord

Imagine an infinite sequence of sentences $s_0, s_1, s_2,
\ldots$ each to the effect that every subsequent sentence is untrue:

(s0)
for all k > 0, sk is untrue,
(s1)
for all k > 1, sk is untrue,
(s2)
for all k > 2, sk is untrue, $\ldots$

Formalizing the sentences with a truth predicate, T, we have that for all natural numbers, n, sn is the sentence $\forall k > n, \neg Ts_k$. Note that each sentence refers to (quantifies over) only sentence later in the sequence. No sentence, therefore, refers to itself, even in an indirect, loop-like, fashion. There seems to be no circularity.

Given this set-up, the argument to contradiction goes as follows. For any n:

$Ts_n \Rightarrow \forall k > n, \neg Ts_k$ (*)
$\Rightarrow \neg Ts_{n+1}$

But:

$Ts_n \Rightarrow \forall k > n, \neg Ts_k$ (*)
$\Rightarrow \forall k > n+1, \neg Ts_k$
$\Rightarrow Ts_{n+1}$

Hence, Tsn entails a contradiction, so $\neg Ts_n$. But n was arbitrary. Hence $\forall k \neg Ts_k$, by Universal [Introduction]. In particular, then, $\forall k > 0, \neg Ts_k$, i.e., s0, and so Ts0. Contradiction (since we have already established $\neg Ts_0$).



 

Selmer Bringsjord
1999-01-19