Ponderable 2.3.3

If s1∈[s2], it means that [s1]⊂[s2] and s1∼s2, s2∼s1 and s2∈[s1], therfore [s2]⊂[s1].

[s1]=[s2].

If s1∉[s2], if ∃t∈[s2]∩[s1], [t]=[s1]=[s2], and s1∈[s2]. It’s impossible.

[s1]≠[s2].

Maison