since then we can give each partner a smallest piece and we are done. The hard case is that they agree. Let's call the piece, that both Bob and Carl view as smallest, X1, and the other two pieces, X2 and X3. Step two. Ask Bob and Carl to mark, on each of the pieces X2 and X3, where the piece has to be cut in order to make it equal to X1. We consider several cases. Case 1. Bob's trims are weaker. I.e, if Bob trims X2 to X2' and X3 to X3', such that both X2' and X3' are for him as small as X1, then Carl thinks X1 is still a smallest piece – weakly smaller than X2' and X3'. Then, the following partial division is envy-free: Carl gets X1; Alice gets the smaller of X2' and X3' (both are smaller than X1 for her); Bob gets the piece not taken by Alice (both are equal to X1 for him). Now we have to divide the trimmings E2 and E3. For each trimming, the following is done: Bob cuts it to three equal pieces. The agents choose pieces in the order: Carl, Alice, Bob. Carl is not envious since he chose first; Bob is not envious since he cut; Alice is not envious since she had a (negative) advantage over Carl: in the first step, Carl took X1, while Alice took a piece that is smaller than X1 by max(E2,E3), while in the last step, Alice took two pieces that are worth at most (E2+E3)/2. Case 2. Carl's trims are weaker. I.e, if Carl trims X2 to X2' and X3 to X3', such that both X2' and X3' are for him as small as X1, then Bob thinks X1 is still a smallest piece – weakly smaller than X2' and X3'. Then, we proceed as in Case 1, with the roles of Bob and Carl switched. Case 3. Bob's trim is weaker in X2, and Carl's trim is weaker in X3. I.e, if Bob trims X2 to X2' which is equal to X1 for him, and Carl trims X3 to X3' which is equal to X1 for him, then: For Carl: X2' >= X1 = X3' For Bob: X3' >= X1 = X2' Then, the following partial division is envy-free: Alice gets the smaller of X2' and X3' (both are smaller than X1 for her); Bob gets