ቐ ଵܶ = [ܥ × ܣ]௦௭() = ܶଵ∗
ଶܶ = [ܥ × ܤ]௦௭() = ܶଶ್∗್
 

$$\Rightarrow$$

ழܥ × ܤ > ܥ × ܣ ⇒ ଶܶ < ଵܶ ሳልልልልሰ
-Fourth situation, (a < b = c):
ቐ ଵܶ = [ܥ × ܣ]௦௭() = ܶଵ∗
ଶܶ = [ܥ × ܤ]௦௭() = ܶଶ್∗್
 
ୀ
ଵܶ(݁ݖ݅ݏ ሳልልልልሰଶܶ(݁ݖ݅ݏ = ()
ܥ × ܤ = ܥ × ܣ ⇒
-Fifth situation, (a < b < c): ቐ ଵܶ = [ܥ × ܣ]௦௭() = ܶଵ∗
ଶܶ = [ܥ × ܤ]௦௭() = ܶଶ∗
 
ୀ
ଵܶ(݁ݖ݅ݏ ሳልልልልሰଶܶ(݁ݖ݅ݏ = ()
ܥ × ܤ = ܥ × ܣ ⇒
So, in all five situations, we proved that:
5.2.6 ܥ × ܤ ≥ ܥ × ܣ ⇒ ܤ ≥ ܣ ݂݅
Hence, the second property of S-norm is satisfied.

## 3. Associativity

(ܥ × ܤ) × ܣ
?ܥ × (ܤ × ܣ)⇔
B × C is a string matrix with each element including three characters. These characters are the elements of B ⊗ C,B and C. Since the elements of B and C are single characters, L(B) = L
ᇱ(B) = B and L(C) =
L
ᇱ(C) = C, which consequently yield L(B × C) = B ⊗
C, and L
ᇱ(B × C) = BC. 

Similar to Monotony condition, by assuming Aୟ∗ୟ, Bୠ∗ୠ and Cୡ∗ୡ as three square matrixes and considering four separate situations, we have:
7
-Third situation, (a < c < b):
-First situation, ( ܽ > ܾ, ܿ ):
The elements of matrix ܣ) × ܤ × ܥ(, contain four characters of matrices A ⊗ L
ᇱ(B ⊗ C, B, C), A, B and C *5.2.7* Based on eq. 5.1.3, we have A ⊗ L
ᇱ(B ⊗ C, B, C) = ((A ⊗ B) ⊗ C ) *5.2.8* Moreover, elements of (ܣ × ܤ × (ܥ contain four characters of matrices
( L((A ⊗ B),A, B) ⊗ L
ᇱ(C) ), A, B and C *5.2.9* The first matrix in eq. 5.2.9 is equal to ((A ⊗ B) ⊗
C ), which is the same as eq. 5.2.8.

-Second situation, ( ܽ < ܾ, ܿ ):
ܤ × ܥ is a combination of three matrixes, B ⊗
C, B and C. Since *size*(A) is less than *size*(B ⊗ C) then :
(ܣ × (ܥ ;ܤ ;ܥ ⊗ ܤ))ܮ = ((ܥ × ܤ) × ܣ)ܮ ܮ ⊗ (ܥ ;ܤ ;ܥ ⊗ ܤ)ܮ ] =[ (ܣ)ᇱ ܣ ⊗ (ܥ ⊗ ܤ) =
Moreover, based on eq. 5.2.5 we have:
ܥ × (ܣ × ܤ) = ܥ × (ܤ × ܣ)
But we should also note that:
ܥ × (ܣ ;ܤ ;ܣ ⊗ ܤ) = (ܥ × (ܣ × ܤ))ܮ ܮ ⊗ (ܣ ;ܤ ;ܣ ⊗ ܤ)ܮ ] =[ (ܥ)ᇱ ܥ ⊗ (ܣ ⊗ ܤ) =
Consequently two equations above are equal.

-Third situation, ( ܿ < ܽ < ܾ ): The size of A is less than size (B ⊗ C) which yields:
ܣ ⊗ (ܥ ⊗ ܤ) = ((ܥ × ܤ) × ܣ)ܮ Moreover, ܥ ⊗ (ܣ ⊗ ܤ) = (ܥ × (ܤ × ܣ))ܮ Hence these two equations are equal.

-Forth situation, ( ܾ < ܽ < ܿ ):
Because the size of matrix C is greater than B, it can be 
:and ,ܤ ⊗ ܥ = (ܥ × ܤ)ܮ that inferred ܣ ⊗ (ܤ ⊗ ܥ) = ((ܥ × ܤ) × ܣ)ܮ In addition the following equation holds:
.ܤ ⊗ (ܣ ⊗ ܥ) = (ܥ × (ܤ × ܣ))ܮ Therefore we have:
5.2.10 ܥ × (ܤ × ܣ) = (ܥ × ܤ) × ܣ We can finally conclude that S-norm Associativity is satisfied for all feasible situations.

4. Neutrality *of zero* To satisfy this property it is needed to introduce *zero*.

If the center component of structure element matrix is 
"1", then it eliminates some part of other matrixes in Extended Thinning. Therefore, it is needed that the structure element matrix does not affect other matrices in Extended Thinning. This matrix must have two properties:
- Its center element must be zero. -It must be smaller than any matrix in order to become structure element.

Considering these properties, the smallest matrix with a "0" element in the center is[0]ଵ∗ଵ, which can serve as neutral zero for S-norm.

Based on eq. 5.2.5, 5.2.6, 5.2.10 and the paragraph above, all four conditions for being an S-norm are proven to be satisfied.

: ()ࢠ = ()ࢠ (II Case Let us assume that Aୟ∗ୟ, Bୠ∗ୠ and Cୡ∗ୡ are three square matrixes. Four conditions are proven under different situations:
1.*Commutativity* ܣ × ܤ = ∗[0] = ܤ × ܣ 2.Monotony