Error in `lastRun()` api docs.

[TheDancingCode]

The lastRun() api docs contain an error. Under usage, it provides the following recipe:

const { src, dest, lastRun, watch } = require('gulp');
const imagemin = require('gulp-imagemin');

function images() {
  return src('src/images/**/*.jpg', { since: lastRun(images) })
    .pipe(imagemin())
    .pipe(dest('build/img/'));
}

function watch() {
  watch('src/images/**/*.jpg', images);
}

exports.watch = watch;

However, running this gives a SyntaxError: Identifier 'watch' has already been declared. You can't name the function watch, since watch is already declared as the gulp command in de destructuring assignment.

[phated]

Thanks @TheDancingCode! I've rewritten it to export a default task.