Moreover, this section is minimal since neither {ж} nor {y} are sections
because, for instance, AC ({ж}) = {0, {ж}} but
Φ (в) = {ц + {в, ω} + {в, {2} .m) + {{>■}. o∙. «}} + {», w}
and
({≈}) = {6} + {0,{z} . {z,w}} + {{r} . {i>,<∕}} + {0. {«}} ,
and hence, C'j'' (0) ≠ C, ,∙!ι'})
Also, {^,w} is a section because AC ({z, w}) = {0, {z} , {z, w}} (notice
that the subset {w} is not an active component of {z, w}) and CfCw's (0),
^,w} and ρAA are an equal to
W + {0,W,M} + {W,⅛Q}} + {0,{t}}∙
Moreover, this section is minimal since neither {z} nor {w} are sections
because, for instance, AC ({w}) = {0, { w } } but
4"1 (0) = {!>} + {0. Ы . {y}} + {0. {z}} + {{r} . {i>,<∕}} + {0. {«}}
and
4”’ (W) = {4 + {0. w. ⅛}} + w + {{и, {∙s,9}} + {0, и,
and hence, C^ (0) ≠ C^ ({w}).
The proof that all other components of the decomposition are also mini-
mal sections is similar and left to the reader.
Now, given a set of agents N, any voting by committees /■': A" → 2λ
will be strategy-proof as long as it satisfies the following properties: (a) by
condition (3) of Theorem 1, W''' = W(" = {¼}} and TV™ = W™ = {{⅛}}
for some ii,i2 ∈ N; (b) by condition (1) of Theorem 1, W('' = TV™; and (c)
by condition (2) of Theorem 1, TVr and TVs are complementary.
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