Is it possible to simplify this IF conditional statement in Verilog? - verilog

Question

I'm was trying to create a binary counter, but when I simplified the IF statement, it stopped to work.
This code works:
if(counter<500000)
counter<=counter+1;
else
counter<=0;
if (counter==0)
if(LEDR<262143)
LEDR <= LEDR+1;
else
LEDR<=0;
this doesn't :
if(counter<500000)
counter<=counter+1;
else
counter<=0;
if (counter==0 && LEDR<262143)
LEDR <= LEDR+1;
else
LEDR<=0;

Answer 1

The two versions of your code are not equivalent.
In your original version, the else part is contained inside the counter == 0 condition, and will be executed when counter == 0 and LEDR >= 262143:
if (counter==0)
if(LEDR<262143)
LEDR <= LEDR+1;
else // counter must be 0 here
LEDR<=0;
In the "simplified" version, the else part will be executed when the opposite of counter == 0 && LEDR < 262143 is true, which is the case if counter != 0 or LEDR >= 262143.
if (counter==0 && LEDR<262143)
LEDR <= LEDR+1;
else // counter could be different from 0
LEDR<=0;
You can visualize the difference by listing all possible combinations in a table:
counter == 0 | LEDR < 262143 | LEDR <= 0 executed | LEDR <= LEDR+1 executed
| | orig. new | orig. new
-------------+---------------+--------------------+------------------------
false | false | no *yes* | no no
false | true | no *yes* | no no
true | false | yes yes | no no
true | true | no no | yes yes
As you can see, the new version behaves differently when counter != 0.
Actually, it is not possible to simplify the desired behaviour to a single if-else statement, because this would mean that you can only distinguish between the two cases of executing either LEDR <= LEDR + 1 or LEDR <= 0. But in the original code there is a third case (when counter != 0) where you execute nothing at all and leave LEDR untouched.

Answer 2

if (counter==0)
if(LEDR<262143)
LEDR <= LEDR+1;
else
LEDR<=0;
versus
if (counter==0 && LEDR<262143)
LEDR <= LEDR+1;
else
LEDR<=0;
The first condition for both is the same but you have altered the condition for the second argument.
First example is if ((counter==0) && !(LEDR<262143))
Second example is if !(counter==0 && LEDR<262143)
Using de-morgans law the second example could be:
if (counter!=0 || !(LEDR<262143) ) or
if (counter!=0 || (LEDR=<262143))
Therefore the if statement is not simplified but functionally different.