Glup V4.0.0 'gulp start' is not a function

As Derek said, in v4 of gulp you have to use gulp.series or gulp.parallel;

Another differences for v4 is that you need to explicit when the task is done, you have 6 ways to do it.. I'll link this answer for a really good explanation Gulp error: The following tasks did not complete: Did you forget to signal async completion?

Mohamed explanation might be useful to understand in depth the reasons, you can find both: Migrate gulp.start function to Gulp v4

I have tried the solution but it needed more code, so I will write my main changes here:

gulp.task("build",
    gulp.series(gulp.parallel("css", "js", "html"), function (done) {
        done();
    })
);


  gulp.task("start", gulp.series("build", "browser-sync"), function (done) {
      devMode = true;
      gulp.watch(["./src/css/**/*.css'"], ["css"]);
      gulp.watch(["./src/js/**/*.js"], ["js"]);
      gulp.watch(["./src/templates/**/*.html"], ["html"]);
   done();
});

in all of the other task, I simply added the done() function ( with the exception of the task html because it has a return that already explicit when the task is done)

Hope this will help someone else, have a nice day :)

gulp.start has been deprecated in v4. Depending on your needs, you can use gulp.series or gulp.parallel instead.

- gulp.task('start', function() {
-   devMode = true;
-   gulp.start(['build', 'browser-sync']);
+ gulp.task('start', gulp.series('build', 'browser-sync'), function(done) {
+   devMode = true;
    gulp.watch(['./src/css/**/*.css'], ['css']);
    gulp.watch(['./src/js/**/*.js'], ['js']);
    gulp.watch(['./src/templates/**/*.html'], ['html']);
  });

This question is probably a duplicated of this one, but since that question hasn't got an accepted answer I'll just echo Mark's answer.