You might be getting confused by your poor use of indentation, and your irregular use of line breaks vs. spaces between statements. Here's how the nested "for" loops in the OP script look when formatted in the "normal" Perlish way (and with a normal use of "print"):

for ( $a = 0; $a <= $i; $a++ ) {
    for ( $s = 0; $s <= $i; $s++ ) {
        my $d = 0;
        print "$a\t$s\n";
        if ( $a != $s ) {
            print "Yes.->$a$s\n";
            last;
        }
        print "system command \$a $a, then for \$s $s".$n;
        $d++;
    }
}
[download]

Proper indentation is important, especially when you're already confused about other things.

... when i need it to do this ...

I'm not really sure where you're going with this, but maybe something like the following can serve as a point of departure:

c:\@Work\Perl\monks>perl -wMstrict -e
"for my $i (0 .. 3) {
   for my $j (0 .. 3) {
     print qq{system command \$i $i, then for \$j $j \n};
     }
   }
"
system command $i 0, then for $j 0
system command $i 0, then for $j 1
system command $i 0, then for $j 2
system command $i 0, then for $j 3
system command $i 1, then for $j 0
system command $i 1, then for $j 1
system command $i 1, then for $j 2
system command $i 1, then for $j 3
system command $i 2, then for $j 0
system command $i 2, then for $j 1
system command $i 2, then for $j 2
system command $i 2, then for $j 3
system command $i 3, then for $j 0
system command $i 3, then for $j 1
system command $i 3, then for $j 2
system command $i 3, then for $j 3
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Update: Then maybe something along the lines of:

c:\@Work\Perl\monks>perl -wMstrict -e
"for my $i (0 .. 3) {
   INNER:
   for my $j (0 .. 3) {
     next INNER if $i == $j;
     print qq{system command \$i $i, then for \$j $j \n};
     }
   }
"
system command $i 0, then for $j 1
system command $i 0, then for $j 2
system command $i 0, then for $j 3
system command $i 1, then for $j 0
system command $i 1, then for $j 2
system command $i 1, then for $j 3
system command $i 2, then for $j 0
system command $i 2, then for $j 1
system command $i 2, then for $j 3
system command $i 3, then for $j 0
system command $i 3, then for $j 1
system command $i 3, then for $j 2
[download]

First off, welcome to the Monastery. We hope you enjoy your stay.

Looking at your post, it is hard to understand what you are trying to achieve. Your specification is not so clear. But lets we what we can do...

my @a1 = qw(My name is Andrew);
my @a2 = qw(My dog named  Andrew);
[download]

First off you define two arrays of words, but later only one is used. Perhaps this is a clue...

foreach my $a1 (@a1){
  $i++;
  #print $a1 . "\n"; 
}
[download]

Is this here to count the words in @a1 and set $i? If so we can skip it, and go an easier way later on

First:for($a = 0; $a <= $i; $a++){
    Second:for($s = 0; $s <= $i; $s++){
        my $d = 0;
        print $a.$t;
        print $s.$n;
        if($a!=$s){
        print "Yes.->".$a.$s.$n; last}
        print "system command \$a $a, then for \$s $s".$n;
        $d++;
    }
}
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OK, so we have two loops, counting from 0 to $i. We have a $d in there that I don't think you use again. Then we print out a line, when the iterators are not equal. All very mysterious.

Wild Guess

Perhaps you are trying to compare the words in the lists? I see two word lists, and some initialisation based on the length of one of them. We may not even need nested loops. If that's what you are after, how about something like this...

use strict;
use warnings;

my @a1 = qw(My name is     Andrew);
my @a2 = qw(My  dog named  Andrew);

my $i = 0;
while ($a1[$i] and $a2[$i]) { # make sure something is in both arrays
    if ($a1[$i] ne $a2[$i]) {
        print "system command \$a1 $a1[$i], then for \$a2 $a2[$i]\n";
    }
    else {
        print "# \$a1 $a1[$i], and \$a2 $a2[$i] are the same\n";
    }
    $i++;
}
[download]

Like I said, a wild guess. Please do update your specification, so we can help a little more

Cheers,
R.

Another approach to a wild guess:

c:\@Work\Perl\monks>perl -wMstrict -e
"my @a1 = qw(My name is    Andrew);
 my @a2 = qw(My dog  named Andrew);
 ;;
 for my $w1 (@a1) {
   INNER:
   for my $w2 (@a2) {
     next INNER if $w1 eq $w2;
     print qq{system command a1 $w1 and a2 $w2 \n};
     }
   }
"
system command a1 My and a2 dog
system command a1 My and a2 named
system command a1 My and a2 Andrew
system command a1 name and a2 My
system command a1 name and a2 dog
system command a1 name and a2 named
system command a1 name and a2 Andrew
system command a1 is and a2 My
system command a1 is and a2 dog
system command a1 is and a2 named
system command a1 is and a2 Andrew
system command a1 Andrew and a2 My
system command a1 Andrew and a2 dog
system command a1 Andrew and a2 named
[download]

Others have commented on the difference between last (found in the OPed code example), which completely exits its associated loop, and next, which terminates the current iteration of its loop and begins the next one (if there is one).

---

Give a man a fish:  <%-(-(-(-<

for my $a (0..$i) {
    for my $s (0..$s) {
        if ($a != $s) {
            # do something
        }
    }
}
[download]