When trying to run gulp sass bash throws a typeerror: gulp.task is not a function. What is the reason?

Question

Windows

The reason was a typo - I assigned var gulp twice, instead of var gulp and var sass.
When trying to run gulp sass through the bash, the following error is generated:

gulp.task('sass', function () {
     ^

TypeError: gulp.task is not a function
    at Object.<anonymous> (C:\Users\Jeddid\Documents\!important\mysassa\sass1\gulpfile.js:6:6)
    at Module._compile (module.js:570:32)
    at Object.Module._extensions..js (module.js:579:10)
    at Module.load (module.js:487:32)
    at tryModuleLoad (module.js:446:12)
    at Function.Module._load (module.js:438:3)
    at Module.require (module.js:497:17)
    at require (internal/module.js:20:19)
    at Liftoff.handleArguments (C:\Users\Jeddid\AppData\Roaming\npm\node_modules\gulp\bin\gulp.js:116:3)
    at Liftoff.<anonymous> (C:\Users\Jeddid\AppData\Roaming\npm\node_modules\gulp\node_modules\liftoff\index.js:198:16)

contents of gulpfile.js:

var gulp = require('gulp');
var gulp =  require('gulp-sass');

gulp.task('sass', function(){
  return gulp.src('source-files')
    .pipe(sass()) // Using gulp-sass
    .pipe(gulp.dest('destination'))
});

What is the reason? Google search did not return results specifically for this type of error in this context

Answer 1

//var gulp =  require('gulp-sass');
var sass =  require('gulp-sass');

Answer 2

.pipe(sass()
close bracket, for example