Example 3.7. Let L = {0, a, b, c} b

Example 3.7. Let L = {0, a, b, c} be a discrete ADL. Define a mapping f : L −→
L by f(x) = 0 for all x ∈ L. Then clearly f is a homomorphism on L. Now
take A = {a, b}. Then clearly A∗ = {0} and f(A) = {0}. Hence f(A∗) = {0} and
{f(A)}∗ = L. Therefore {f(A)}∗ 6= f(A∗).

ผลลัพธ์ (อังกฤษ) 2:

Example 3.7. Let L = {0, a, b, c} be a discrete ADL. Define a Mapping F: L - →
L by F (x) = 0 for all x ∈ L. Then F is clearly a Homomorphism on L. Now
Take A = {a, B}. Then clearly A * = {0} and f (A) = {0}. Hence F (A *) = {0} and
{F (A) = L. Therefore {F} * (A)} * 6 = F (A *).

ผลลัพธ์ (อังกฤษ) 3:

Example 3.7. Let L = {,,, 0 a B C} be a discrete ADL. Define a mapping f: L − - >
L by F (x) = 0 for all x ∈ L. Then clearly. F is a homomorphism on L. Now
take A = {,}. Then a B clearly A ∗ = {0} and f (A) = {0}. Hence f (A ∗) = {}
{f 0 and (A)} = L, ∗. Therefore {f (A)} ∗ 6 = f (A ∗).