, (a ? 2, 4) ? C. Consequently a ? 2 sends 1/4 to a ? 1 (by Rule 1) satisfying it and sends 1/2 to a by Rule 2. Finally
, then (a ? 2, 1) ? C or (a ? 2, 4) ? C. If (a ? 2, 2) ? C, then a ? 2 sends 1, vol.2
, (a ? 1, 3)}. Then a ? 1 sends 3/4 to a by Rule 1, so exc 1 (a) = ?1/2. Since C
, Thus (a ? 2, 1), (a ? 2, 4) ? C. In particular, a ? 2 sends 3/4 to a ? 1 by Rule 1, so exc 1 (a ? 1) = ?1/2. Moreover, |C ? Q a?3 | ? 1, for otherwise a ? 3 sends 1/2 to a ? 1 by Rule 2 and 1/4 to a by Rule 3 and a is satisfied
Thus (a ? 2, 1), (a ? 2, 2), (a ? 2, 4) ? C. Moreover |C ? Q a?3 | ? 1 for otherwise a ? 3 sends 1/4 to a ? 2, a ? 1, and a by Rules 1, 2 and 3, respectively, and a is satisfied ,
Thus, (a ? 2, 1), (a ? 2, 3), (a ? 2, 4) ? C. Moreover |C ? Q a?3 | ? 1 for otherwise a ? 3 sends 1/4 to a ? 2, a ? 1, and a by Rules 1, 2 and 3, respectively, and a is satisfied ,
, so either (a ? 2, 1) ? C or (a ? 2, 4) ? C. But |C ? Q a?2 | ? 2 for otherwise a ? 2 sends 5/4 to a ? 1 by Rule 1 and 1/4 to a by Rule 2, and a is satisfied, vol.2
,
, so either (a?2, 1) ? C or (a?2, 4) ? C. But |C ?Q a?2 | ? 1 for otherwise a?2 sends 1/4 to a?1 by Rule 1 and 1/4 to a by Rule 2, and a is satisfied
, , vol.2
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