Is it possible to stop pipe execution for a specific file in gulp?

Question

JavaScript

Hello! There is such a gallp task:

function jsBuild () {
  return gulp.src( path.app.js )
    .pipe( sourcemaps.init() )
    .pipe( babel() )
    .pipe( concat( 'main.js' ) )
    .pipe( uglify( {
      toplevel: true
    } ) )
    .pipe( replace( '"use strict";', '' ) )
    .pipe( sourcemaps.write( './' ) )
    .pipe( gulp.dest( path.dist.js ) )
    .pipe( browserSync.stream() );
}

and these are the paths:

app: {
    html: 'app/*.html',
    js: [
      'app/js/jquery-3.4.1.min.js',
      'app/js/jquery.treeSelector.js',
      'app/js/parallax_background.js',
      'app/js/TweenLite.min.js',
      'app/js/wow.min.js',
      'app/js/CSSPlugin.min.js',
      'app/js/popper.min.js',
      'app/js/main.js',
    ],
    mainJs: 'app/js/main.js',
    scss: 'app/css/main.scss',
    img: 'app/img/**/*.*',
    fonts: 'app/fonts/**/*.*'
  },

is it possible to somehow make it so that when the task is done for popper.js, not all pipes are executed? Popper breaks after beable (.babelrc:

{
    "presets": [
        "@babel/env"
    ]
}

)
It turns out that you need to skip the pipe with the beyble / babel, whatever you like) and with uglify, because uglify breaks when it sees const or God forbid es modules.
Thanks