In this paper, we investigate the following problem: give a quasi-Boolean function Ψ(x
1, …, x
n
) = (a ∧ C) ∨ (a
1 ∧ C
1) ∨ … ∨ (a
p
∧ C
p
), the term (a ∧ C) can be deleted from Ψ(x
1, …, x
n
)? i.e., (a ∧ C) ∨ (a
1 ∧ C
1) ∨ … ∨ (a
p
∧ C
p
) = (a
1 ∧ C
1) ∨ … ∨ (a
p
∧ C
p
)? When a = 1: we divide our discussion into two cases. (1) ℑ1(Ψ,C) = ø, C can not be deleted; ℑ1(Ψ,C) ≠ ø, if S
i
0 ≠ ø (1 ≤ i ≤ q), then C can not be deleted, otherwise C can be deleted. When a = m: we prove the following results: (m∧C)∨(a
1∧C
1)∨…∨(a
p
∧C
p
) = (a
1∧C
1)∨…∨(a
p
∧C
p
) ⇔ (m ∧ C) ∨ C
1 ∨ … ∨C
p
= C
1 ∨ … ∨C
p
. Two possible cases are listed as follows, (1) ℑ2(Ψ,C) = ø, the term (m∧C) can not be deleted; (2) ℑ2(Ψ,C) ≠ ø, if (∃i
0) such that S¢i0 S'_{i_0 } = ø, then (m∧C) can be deleted, otherwise ((m∧C)∨C
1∨…∨C
q
)(v
1, …, v
n
) = (C
1 ∨ … ∨ C
q
)(v
1, …, v
n
)(∀(v
1, …, v
n
) ∈ L
3
n
) ⇔ (C
1′ ∨ … ∨ C
q
′)(u
1, …, u
q
) = 1(∀(u
1, …, u
q
) ∈ B
2
n
).
KeywordsLattice–Boolean function–Quasi-Boolean algebra–Quasi-Boolean function