Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$. By definition, we have that there exist injections $f: S \to T$ and $g: T \to S$. We are going to try to build a sequence $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows. Consider any $t_1 \in T$. By Law of Excluded Middle there are two choices: {{begin-eqn}} {{eqn | l = \exists s_1 \in S: \map f {s_1}       | r = t_1       | c =  }} {{eqn | l = \neg \exists s_1 \in S: \map f {s_1}       | r = t_1       | c =  }} {{end-eqn}} Suppose $\exists s_1 \in S: \map f {s_1} = t_1$. Because $f$ is injective, such an $s_1$ is unique. So we can choose $s_1 = \map{f^{-1} } {t_1}$. Again, by Law of Excluded Middle there are two further choices: {{begin-eqn}} {{eqn | l = \exists t_2 \in T: \map g {t_2}       | r = s_1       | c =  }} {{eqn | l = \neg \exists t_2 \in T: \map g {t_2}       | r = s_1       | c =  }} {{end-eqn}} Suppose $\exists t_2 \in T: \map g {t_2} = s_1$. Because $f$ is injective, such an $t_2$ is unique. Similarly, we choose $s_2 = \map {f^{-1} } {t_2}$, if it exists. This process goes on until one of the following happens: :We reach some $s_n \in S$ such that $\neg \exists t \in T: \map g t = s_n$. This may be possible because $g$ may not be a surjection. :We reach some $t_n \in T$ such that $\neg \exists s \in S: \map f s = t_n$.This may be possible because $f$ may not be a surjection. :The process goes on for ever. For each $t \in T$, then, there is a well-defined process which turns out in one of the above three ways. We partition $T$ up into three subsets that are mutually disjoint: : Let $T_A = \{$ all $t \in T$ such that the process ends with some $s_n\}$ : Let $T_B = \{$ all $t \in T$ such that the process ends with some $t_n\}$ : Let $T_C = \{$ all $t \in T$ such that the process goes on for ever$\}$. We can do exactly the same thing with the elements of $S$: : Let $S_A = \{$ all $s \in S$ such that the process ends with some $s_n\}$ : Let $S_B = \{$ all $s \in S$ such that the process ends with some $t_n\}$ : Let $S_C = \{$ all $s \in S$ such that the process goes on for ever$\}$. What we need to do is show that $S \sim T$. We do this by showing that $S_A \sim T_A$, $S_B \sim T_B$ and $S_C \sim T_C$. * The restriction of $f$ to $S_A$ is a bijection from $S_A$ to $T_A$. To do this we need to show that: # $s \in S_A \implies \map f s \in T_A$; # $\forall t \in T_A: \exists s \in S_A: \map f s = t$. Let $s \in S_A$. Then the process applied to $s$ ends in $S$. Now consider the process applied to $\map f s$. Its first step leads us back to $s$. Then it continues the process, applied to $s$, and ends up in $S$. Thus $\map f s \in T_A$. Thus $s \in S_A \implies \map f s \in T_A$. Now suppose $t \in T_A$. Then the process applied to $t$ ends in $S$. In particular, it must have a first stage, otherwise it would end in $T$ with $t$ itself. Hence $t = \map f s$ for some $s$. But the process applied to this $s$ is the same as the continuation of the process applied to $t$, and therefore it ends in $S$. Thus $s \in S_A$ as required. Hence the restriction of $f$ to $S_A$ is a bijection from $S_A$ to $T_A$. We can use the same argument to show that $g: T_B \to S_B$ is also a bijection. Hence $g^{-1}: S_B \to T_B$ is a bijection. Finally, suppose $t \in T_C$. Because $f$ is an injection, $t = \map f s$ for some $s$, and the process applied to $t$ must start. And this $s$ must belong to $S_C$, because the process starting from $s$ is the same as the process starting from $t$ after the first step. This never ends, as $t \in T_C$. Now we can define a bijection $h: S \to T$ as follows: :$\map h x = \begin{cases} \map f s: x \in S_A \\ \map f s: x \in S_C \\ \map {g^{-1} } x: x \in S_B \end{cases}$ The fact that $h$ is a bijection follows from the facts that: :$(1): \quad S_A$, $S_B$ and $S_C$ are mutually disjoint :$(2): \quad T_A$, $T_B$ and $T_C$ are mutually disjoint :$(3): \quad f$, $f$ and $g^{-1}$ are the bijections which respectively do the mappings between them. {{Qed}}