Number that can eat itself Given a positive integer, output a truthy/falsy value as to whether the number can eat itself.

Rules

Leftmost is the head, rightmost is the tail

If the head is greater than or equal to the tail, the head eats the tail and the new head becomes their sum.

If \$sum \ge 10 \$ then the head is replaced by \$sum \mod 10\$.

\$sum=0\$ cannot be ignored, however the input number will never have any leading zeroes.

Example:

number=2632
head-2, tail-2

2632 -&gt; 463
head-4, tail-3

463 -&gt; 76
head-7, tail-6

76 -&gt; 3
If only one digit remains in the end, the number can eat itself.


If at any point the head cannot eat the tail, the answer will be False.

number=6724
072
False (0&lt;2)


Test Cases:

True:
[2632, 92258, 60282, 38410,3210, 2302, 2742, 8628, 6793, 1, 2, 10, 100, 55, 121]

False:
[6724, 47, 472, 60247, 33265, 79350, 83147, 93101, 57088, 69513, 62738, 54754, 23931, 7164, 5289, 3435, 3949, 8630, 5018, 6715, 340, 2194]


This is code-golf so shortest code wins.
JavaScript (ES6),  52 51  50 bytes
Saved 1 byte thanks to @tsh
Takes input as a string. Returns a Boolean value.

f=n=&gt;n&gt;[n%10]?f(-(-n[0]-n)%10+n.slice(1,-1)):!n[1]

Try it online!
Commented
f = n =&gt;                 // f = recursive function taking n (a string)
  n &gt; [n % 10]           // The last digit is isolated with n % 10 and turned into a
                         // singleton array, which is eventually coerced to a string
                         // when the comparison occurs.
                         // So we do a lexicographical comparison between n and its
                         // last digit (e.g. '231'&gt;'1' and '202'&gt;'2', but '213'&lt;'3').
  ?                      // If the above result is true:
    f(                   //   do a recursive call:
      -(-n[0] - n) % 10  //     We compute (int(first_digit) + int(n)) mod 10. There's no
                         //     need to isolate the last digit since we do a mod 10 anyway.
      + n.slice(1, -1)   //     We add the middle part, as a string. It may be empty.
    )                    //   end of recursive call
  :                      // else:
    !n[1]                //   return true if n has only 1 digit, or false otherwise