What do you think of my "Battle Resolution System v1.1,1.2,1.3"?

I'm Back with some few Ideas here...

If you can still remember my old topic here it is: (Forum)
http://www.bgdf.com/node/2192

Complete Summary: (Game Journal)
http://www.bgdf.com/node/2210

I'm sorry If you could not understand my language (Mathematical representation).
But this is the only way I can explain it clearly...
Tell me what you think, which is the best?

==========Battle Resolution System v1.1==========

Situation 1: If R[A] is > R[B] -----> ( R[A] + N[A] ) - ( R[B] + N[B] ) = C (Original Equation v1.1)
Situation 2: If R[B] is > R[A] -----> ( R[B] + N[B] ) - ( R[A] + N[A] ) = C (Original Equation v1.1)

Equation 1: B-C=D (Winner A)
Equation 2: A-C=D (Winner A)
Equation 3: RA = RB (Draw)
---------------------A&B -1 = ( If R=1 or 2 )
---------------------A&B -2 = ( If R=3 or 4 )
---------------------A&B -3 = ( If R=5 or 6 )

A= Side A (Max. of 9 units)
B= Side B (Max. of 9 units)
C= Difference between R[A] and R[B] or vice versa (Damage: 1-5)
D= Result (Units Left)
R= Die Roll (Standard Die: 1-6)
N= No. of Units
R[A]= Side A Die Roll
R[B]= Side B Die Roll
N[A]= No. of Units in Side A
N[B]= No. of Units in Side A

Notes:
N[A] & N[B] (Added to increase Chances)

==========Battle Resolution System v1.2==========

Just remove N[A] & N[B] in the Equations: (Equal chances for both sides)

Situation 1: If R[A] is > R[B] -----> ( R[A] + N[A] ) - ( R[B] + N[B] ) = C (Original Equation v1.1)
Situation 2: If R[B] is > R[A] -----> ( R[B] + N[B] ) - ( R[A] + N[A] ) = C (Original Equation v1.1)

Situation 1: If R[A] is > R[B] -----> R[A] - R[B] = C (Modified Equation v1.2)
Situation 2: If R[B] is > R[A] -----> R[B] - R[A] = C (Modified Equation v1.2)

Examples:

(Winner A)
Situation 1:
A= 5 --> 5
B= 9 --> 8
Situation 2:
A= 7 --> 7
B= 7 --> 6
Situation 3:
A= 9 --> 9
B= 5 --> 4

(Winner B)
Situation 4:
A= 5 --> 4
B= 9 --> 9
Situation 5:
A= 7 --> 6
B= 7 --> 7
Situation 6:
A= 9 --> 8
B= 5 --> 5

(Draw)
Situation 7:
A= 5 --> 4,3,2
B= 9 --> 8,7,6
Situation 8:
A= 7 --> 6,5,4
B= 7 --> 6,5,4
Situation 9:
A= 9 --> 8,7,6
B= 5 --> 4,3,2

For Situation 1-6 (Assume C=1)
For Situation 7-9 (C[1]=1, C[2]=2, C[3]=3)

==========Battle Resolution System v1.3==========

Just Replace N[A] & N[B] with S:

Situation 1: If R[A] is > R[B] -----> ( R[A] + N[A] ) - ( R[B] + N[B] ) = C (Original Equation v1.1)
Situation 2: If R[B] is > R[A] -----> ( R[B] + N[B] ) - ( R[A] + N[A] ) = C (Original Equation v1.1)

Situation 1: If R[A] is > R[B] -----> R[A] - R[B] = C (Modified Equation v1.3)
Situation 2: If R[B] is > R[A] -----> R[B] - R[A] = C (Modified Equation v1.3)

S= Bonus (Will depend on the N of A&B)

S= 1 (3 units)
S= 2 (6 units)
S= 3 (9 units)

If N=1 -----> S= 0
If N=2 -----> S= 0
If N=3 -----> S= 1
If N=4 -----> S= 1
If N=5 -----> S= 1
If N=6 -----> S= 2
If N=7 -----> S= 2
If N=8 -----> S= 2
If N=9 -----> S= 3