Modernising COBOL takes time and care (opens in new tab)

Hey Grok, Can you write the simplest recursive fibonacci-generator in COBOL?

       IDENTIFICATION DIVISION.
       PROGRAM-ID. FIBONACCI.
       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01  NUM         PIC 9(10).
       01  RESULT      PIC 9(10).
       PROCEDURE DIVISION.
       MAIN-PARA.
           MOVE 10 TO NUM
           CALL 'FIB' USING NUM RESULT
           DISPLAY 'FIBONACCI OF ' NUM ' IS ' RESULT
           STOP RUN.
       IDENTIFICATION DIVISION.
       PROGRAM-ID. FIB.
       DATA DIVISION.
       LINKAGE SECTION.
       01  N           PIC 9(10).
       01  FIB-RESULT  PIC 9(10).
       PROCEDURE DIVISION USING N FIB-RESULT.
           IF N <= 1
               MOVE N TO FIB-RESULT
           ELSE
               COMPUTE FIB-RESULT = FUNCTION FIB(N - 1) + FUNCTION FIB(N - 2)
           END-IF
           GOBACK.
       END PROGRAM FIB.
       END PROGRAM FIBONACCI.