Let P be a partially ordered set. We prove that if n is sufficiently large, then there exists a packing P of copies of P in the Boolean lattice (2 [n] , ⊂) that covers almost every element of 2 [n] : P might not cover the minimum and maximum of 2 [n] , and at most |P | − 1 additional points due to divisibility. In particular, if |P | divides 2 n − 2, then the truncated Boolean lattice 2 [n] − {∅, [n]} can be partitioned into copies of P . 1 otherwise, there is no copy of P covering [n] or ∅ in 2 [n] 1 The second conjecture of Lonc is concerned with posets P which do not necessarily satisfy that the size of P is a power of 2, or have a unique minimum and maximum. In this case, it is still reasonable to believe that there exists a P -packing in 2 [n] that covers almost everything. More precisely, the conjecture states that if n is sufficiently large and 2 n − 2 is divisible by |P |, then the truncated Boolean lattice 2 [n] − {∅, [n]} has a P -partition. This conjecture was verified by Lonc [12] in the case P is an antichain, or |P | ≤ 4. Also, Gruslys, Leader and Tomon [7] proposed a relaxation of this conjecture. That is, there exists a constant c(P ) such that 2 [n] has a P -packing that covers all but at most c(P ) elements of 2 [n] for every n. This conjecture was verified by the author of this paper [16] in case P has a unique minimum and maximum (but size not necessarily a power of 2).We settle both of the aforementioned conjectures in the following theorem.Theorem 2. Let P be a poset. There exists n 0 = n 0 (P ) such that if n ≥ n 0 , then there exists a P -packing P of 2 [n] − {∅, [n]} such that the number of elements not covered by P is at most |P | − 1.Our paper is organized as follows. In the next subsections, we discuss some related partitioning results and we introduce our notation. In Section 2, we prove Theorem 1. In Section 3, we prove Theorem 2. We finish our paper with some remarks and open problems in Section 4.