Please convert the below content to rendered equations without any changes and errors.This circuit is a **2-bit magnitude comparator** implemented using logic gates. Outputs required: * (A < B) * (A > B) * (A = B) The small blocks on the left compute **bit equality or inequality signals** for each bit pair. Let the inputs be: [ A = A_1A_0,\qquad B = B_1B_0 ] From the image the inputs are: | Bit | A | B | | --- | - | - | | MSB | 1 | 1 | | LSB | 0 | 1 | So [ A = 10_2 = 2 ] [ B = 11_2 = 3 ] --- # Step 1: MSB comparison ### Inverters [ A_1' = 0 ] [ B_1' = 0 ] ### AND gates Top AND: [ A_1 B_1' = 1 \cdot 0 = 0 ] Bottom AND: [ A_1' B_1 = 0 \cdot 1 = 0 ] ### OR gate (node (x_3)) [ x_3 = 0 + 0 = 0 ] So [ \boxed{x_3 = 0} ] This means **MSBs are equal**. --- # Step 2: Next block Inputs shown: [ A = 0, \quad B = 0 ] ### Inverters [ A' = 1 ] [ B' = 1 ] ### AND gates [ A B' = 0 \cdot 1 = 0 ] [ A' B = 1 \cdot 0 = 0 ] ### OR gate (node (x_2)) [ x_2 = 0 + 0 = 0 ] [ \boxed{x_2 = 0} ] --- # Step 3: Third block Inputs: [ A = 1,\quad B = 0 ] ### Inverters [ A' = 0 ] [ B' = 1 ] ### AND gates [ A B' = 1 \cdot 1 = 1 ] [ A' B = 0 \cdot 0 = 0 ] ### OR gate (node (x_1)) [ x_1 = 1 + 0 = 1 ] [ \boxed{x_1 = 1} ] --- # Step 4: Last block Inputs: [ A = 0,\quad B = 1 ] ### Inverters [ A' = 1 ] [ B' = 0 ] ### AND gates [ A B' = 0 ] [ A' B = 1 ] ### OR gate (node (x_0)) [ x_0 = 1 ] [ \boxed{x_0 = 1} ] --- # Step 5: Final outputs Because the **MSBs are equal**, the comparison depends on the LSB. [ A_0 = 0,\quad B_0 = 1 ] Therefore [ A < B ] --- # Final Results ### Internal nodes [ x_3 = 0 ] [ x_2 = 0 ] [ x_1 = 1 ] [ x_0 = 1 ] --- ### Comparator outputs [ \boxed{A < B = 1} ] [ \boxed{A > B = 0} ] [ \boxed{A = B = 0} ] --- ✅ Final interpretation: [ A = 2,\quad B = 3 ] [ \boxed{A < B} ] --- If you want, I can also show a **super quick 10-second trick to solve magnitude comparator problems in exams without analyzing every gate.**