ܥ × ܤ ≥ ܥ × ܣ ⇒ ܾ = ܽ ݂݅
Assuming $Max(a,c)=d$:  $$\begin{array}{l}\left(A\times C=[T_{1}]_{d\times d}=T_{1_{d\times d}}\right.\\ \left.B\times C=[T_{2}]_{d\times d}=T_{2_{d\times d}}\right.\\ \left.a=b\right.\\ \left.T_{1}=T_{2}\right.\\ \left.\Rightarrow\right.\left.A\times C=\right.\left.B\times C\right.\end{array}$$
3. *Associativity* Since the size of matrix A is equal to B, we analyze three situations for different sizes of matrix C:
-First situation, (ܽ = ܾ > ܿ) :
In this situation, (ܤ × ܥ (contains three matrixes B ⊗ C, B, and C with size b*b. Therefore ܮ ⊗ (ܣ)ܮ = (ܥ × ܤ) × ܣ∗[0] = (ܥ × ܤ)ᇱ This is because the size of two matrices Aa*a and Bb*b are the same (a=b).

-Second situation, (ܽ = ܾ = ܿ):
Because all the three matrices have the same size, the Extended Thinning for each of them with respect to another is [0]∗, and consequently the final result of Extended Thinning for three matrixes is[0]∗.

-Third situation, (ܽ = ܾ < ܿ):
The size of matrix B is smaller than C, and (ܤ× 
ܥ (consists of C ⊗ B, C and B, with a size of c*c.

ܮ ⊗ (ܥ × ܤ)ܮ = ൯)ܥ × ܤ) × ܣ൫ܮ(ܣ)ᇱ ܣ ⊗ (ܤ ⊗ ܥ) =
Moreover, ܣ × ܤ contains [0]∗,ܣ, and ܤ with a size of a*a. 

ܮ ⊗ (ܥ)ܮ = ൯ܥ × (ܤ × ܣ)൫ܮ(ܤ × ܣ)ᇱ ܮ ⊗ ܥ =ܤ ⊗ (ܣ ⊗ ܥ) = (ܤ ,ܣ,[0])ᇱ These matrixes are equal. So, Associativity is satisfied.

4. Neutrality *of zero* Similar to Case I, it can be proven that the neutral matrix is [0]ଵ∗ଵ.

Therefore, all four conditions are satisfied in all feasible situations and Extended Thinning can consequently serve as an S-norm.

## 5.3. Extended Thickening

Let us assume that A and B are square matrixes. The Extended Thickening operator is defined as:

$$A\circ B$$
$$\mathbf{\Sigma}=$$
$$[\mathbf{1}]_{s i z e(A)}\qquad\qquad\qquad i f\ s i z e(A)$$

⎩
⎪ ⎨ ⎪
⎧()ࢠ[]()ࢠ = ()ࢠ ൫ ⊙ ൯)) ,()൫ ], ൯))′ ,()ᇱ
 [ ൯))′ ,()′൫ 5.3.1 All of the definitions of Max, Min, Save, L, L' and R 
are exactly the same as before. Hence, the proof of Commutativity, Monotony and Associativity would be the same as before. It is enough to introduce Neutrality of one, the fourth property of T-norm, to prove that this new operator can serve as a T-norm.

If the center element of structure element matrix is "0", 
then the Extended Thickening adds structure element data to the main matrix. Therefore, it is needed that the structure element matrix does not affect other matrices in Extended Thickening. This matrix must have two properties:
- Its center element must be one.

-It must be smaller than any matrix in order to become structure element.

Considering these properties, the smallest matrix with 
"1" in the center is [1]ଵ∗ଵ, which is the neutral element for a T-norm.

Therefore, we proposed two new operators that can serve as fuzzy S-norm and T-norm.

Next section proves the duality of these operators.

## 5.4.Extended Thinning, Extended Thickening And Demorgans Law

Let us assume that A and B are two square matrixes.

Without affecting the generality of our proof, we assume that size of A is greater than B:
(ܤ)݁ݖ݅ݏ < (ܣ)݁ݖܵ݅ ⟹ ܤ < ܣ ݂݅

$$\begin{array}{r l}{A^{c}}&{{}\operatorname{size}(A)>\operatorname{size}(B)}\\ {\implies}&{{}s i z e(A^{c})>s i z e(B^{c})}\\ {\implies}&{{}A^{c}>B^{c}}&{{}5.4.1}\end{array}$$
$$\frac{5.4.1}{\Longrightarrow}$$
ሳልልልልሰ ቐ
$$\begin{array}{l}{{\left\{\begin{array}{l l}{s i z e(\,M a x(A,B)\,)=S i z e(\,M a x(A^{c},B^{c})\,)}\\ {}\\ {s i z e(\,m i n(A,B)\,)=S i z e(\,m i n(A^{c},B^{c})\,)}\\ {}\\ {5.4.2}\end{array}\right.}}\end{array}$$

Considering equations 5.4.1 and 5.4.2 simplifies proving the duality of operators ܣ × ܤ and ܣ ∘ ܤ to proving the duality of ܣ ⊗ ܤ and ܣ ⊙ ܤ. This proof is previously explained in section 4.3 completely.

## 6. Results

The first results are obtained on a simple data plane applying both original operators of ALM.

IDS and COG are executed in 5.1. Drawbacks of COG 
are then shown from viewpoint of geometry. 

Extended Thickening and Extended Thinning are executed in 5.2 and it is shown that they do not have disadvantages of COG.