Disable functions/aliases in a sourced script - bash

Question

I know I can run an "original" command (not alias) using either \ or "":
\ls
"ls"
This doesn't work for functions though. Also it requires me to use that syntax every time.
Is it possible in a sourced script to disable all functions/aliases from the parent process (one which runs my script)? I.e. if a user in their terminal has some aliases functions defined I want them disabled in my script (but of course I still want to be able to define and use aliases/functions of my own).

Answer 1

Types of Commands in Bash
Bash knows different types of commands which can shadow each other. The precedence of these types is:
aliases
can be defined by the user using alias cmd=...
functions
can be defined by the user using cmd() { ... }
built-ins
are directly implement in bash and cannot be altered. help and enable list all built-ins.
Executable files in $PATH
Meaning if you type cmd arg1 arg2 ... you use the alias cmd if it is defined, otherwise you use the function cmd if it is defined, otherwise you use the built-in cmd if it is built-in, otherwise you use the first executable cmd from the directories in $PATH if there is one, otherwise you end up with the error -bash: cmd command not found.
Which of these cases applies for cmd can be checked using type -a cmd.
Manual Precedence Control
Bash allows you to influence which type to pick using quoting and the built-ins command and builtin.
\cmd
suppresses aliases
uses functions, built-ins, executables
command cmd
suppresses aliases and functions
uses built-ins and executables
builtin cmd
supresses aliases, functions, and executables
uses only built-ins
enable -n cmd
disables the built-in cmd completely, such that afterwards only
aliases, functions, and executables are used
env cmd
not a bash built-in, therefore it doesn't really suppress anything but
uses only executables
Examples
Shadowing is perfectly normal. For instance, bash has its own built-in echo, but your system also has /bin/echo. Both implementations may differ. For instance, my echo from bash 5 supports \uXXXX but my echo from GNU coreutils 8.3 does not. The possibility of such differences becomes even more clear if you add your own implementations using aliases and functions. Here's an example in an interactive bash session ($ is the prompt):
$ echo() { printf "function echo: %s\n" "$*"; }
$ alias echo='printf "alias echo: %s %s %s\n"'
$ type -a echo
echo is aliased to `printf "alias echo: %s %s %s\n"'
echo is a function
echo ()
{
printf "function echo: %s\n" "$*"
}
echo is a shell builtin
echo is /bin/echo
$ echo -e '\u2261'
alias echo: -e \u2261
$ \echo -e '\u2261'
function echo: -e \u2261
# use the built-in (or executable file if there was no such built-in)
$ command echo -e '\u2261'
≡
$ builtin echo -e '\u2261'
≡
# use the executable /bin/echo
$ env echo -e '\u2261'
\u2261
$ enable -n echo
# use the executable /bin/echo (`command` is needed to skip the alias and function)
$ command echo -e '\u2261'
\u2261
Answering your Question
Unfortunately I'm not aware of something like enable to permanently disable alias and function lookup. You could try some hacks like backing up all aliases and functions, doing unset -f and unalias on them, and restoring them at the end. However, unset may fail for readonly functions. The better way would be to use bash -c '... functions and aliases have no effect here ...' for the parts where you don't really need the benefits of source. For the other parts, prefix everything with command.
Please note: The caller who sources your script may even disable or shadow command, builtin, and so on -- therefore you can never be sure that you are actually using the commands you expected. Even writing /usr/bin/env executable or /path/to/the/executable does not help as a function can have the name and $PATH or the file system can be altered.
However, that shouldn't be your concern. The one who sources your script should be responsible for providing the correct environment.

Answer 2

Edit: this answer might no longer be relevant since you edited the question to clarify that the script is being sourced, not being executed in a subshell.
This happens by default. Proof:
$ function x() { echo 'hi'; }
$ x
hi
$ bash
# We are now in a subshell.
$ x
bash: x: command not found
Functions are often defined in one of the shell's startup files: .bashrc, .profile or .bash_profile. Which of these are sourced depends on whether the shell is a login shell and/or an interactive shell. A shell that invoked to execute a shell script is neither a login shell nor an interactive shell, and in this case none of those files are sourced.

Answer 3

EDIT: I should read more carefully, as you don't want to source a script, but be sourced, the following is for the other way around:
Functions
If you source your parent script at the beginning, you can just loop through the defined functions and unset them.
declare -F will list all defined functions but in the format declare -f functioname, so you have to get only the name:
IFS=$'\n'
for f in $(declare -F|cut -d ' ' -f 3); do
unset -f $f
done
Aliases
Alias should not be sourced in as i remember, but if they are there you can do
unalias -a
to unset them all.