# SUBTRACT\_YEARS

### Description

`SUBTRACT_YEARS` takes a date, from either an information or in [ISO8601 format](https://docs.atfinity.io/rule-language/operators/date-operators),  and a number of years to subtract, and returns a date with the given number of years subtracted.

In the special case that the new year doesn't contain the original date (e.g. when subtracting a year from the leap day 29.02.2016) the new date will have the same number of days from the beginning of the year (e.g. 01.03.2015 in the earlier example).

### Example: Date

```
SUBTRACT_YEARS(NOW(), 18)
```

This expression returns the current date 18 years ago.

```
SUBTRACT_YEARS('2000-01-01, 18)
```

This returns the 1st January 1982

### Example: Date with time

If you specify not just a date, but a date and time in ISO8601 format, the function includes the time in the calculation.

```
SUBTRACT_YEARS('2000-01-01T09:35:23', 18)
```

This returns also the 1st January 1982, but at 9:35:23, so `1982-01-01T09:35:23`.