How to tell if one regular expression matches a subset of another regular expression?

I think — in theory — to tell whether regexp A matches a subset of what regexp B matches, an algorithm could:

Compute the minimal Deterministic Finite Automaton of B and also of the "union" A|B.
Check if the two DFAs are identical. This is true if and only if A matches a subset of what B matches.

However, it would likely be a major project to do this in practice. There are explanations such as Constructing a minimum-state DFA from a Regular Expression but they only tend to consider mathematically pure regexps. You would also have to handle the extensions that Python adds for convenience. Moreover, if any of the extensions cause the language to be non-regular (I am not sure if this is the case) you might not be able to handle those ones.

But what are you trying to do? Perhaps there's an easier approach...?

Verification of the post by "antinome" using two regex : 55* and 5* :

REGEX_A: 55* [This matches "5", "55", "555" etc. and does NOT match "4" , "54" etc]

REGEX_B: 5* [This matches "", "5" "55", "555" etc. and does NOT match "4" , "54" etc]

[Here we've assumed that 55* is not implicitly .55.* and 5* is not .5.* - This is why 5* does not match 4]

REGEX_A can have an NFA as below:

  {A}--5-->{B}--epsilon-->{C}--5-->{D}--epsilon-->{E}
           {B} -----------------epsilon --------> {E} 
                          {C} <--- epsilon ------ {E}

REGEX_B can have an NFA as below:

  {A}--epsilon-->{B}--5-->{C}--epsilon-->{D}
  {A} --------------epsilon -----------> {D} 
                 {B} <--- epsilon ------ {D}

Now we can derive NFA * DFA of (REGEX_A|REGEX_B) as below:

  NFA:
  {state A}  ---epsilon --> {state B} ---5--> {state C} ---5--> {state D}
                                              {state C} ---epsilon --> {state D} 
                                              {state C} <---epsilon -- {state D}
  {state A}  ---epsilon --> {state E} ---5--> {state F}
                            {state E} ---epsilon --> {state F} 
                            {state E} <---epsilon -- {state F}

  NFA -> DFA:

       |   5          |  epsilon*
   ----+--------------+--------
    A  |  B,C,E,F,G   |   A,C,E,F
    B  |  C,D,E,F     |   B,C,E,F
    c  |  C,D,E,F     |   C
    D  |  C,D,E,F,G   |   C,D,E,F
    E  |  C,D,E,F,G   |   C,E,F
    F  |  C,E,F,G     |   F
    G  |  C,D,E,G     |   C,E,F,G

                    5(epsilon*)
    -------------+---------------------
              A  |  B,C,E,F,G 
      B,C,E,F,G  |  C,D,E,F,G 
      C,D,E,F,G  |  C,D,E,F,G 

    Finally the DFA for (REGEX_A|REGEX_B) is:
         {A}--5--->{B,C,E,F,G}--5--->{C,D,E,F,G}
                                     {C,D,E,F,G}---5--> {C,D,E,F,G}

         Note: {A} is start state and {C,D,E,F,G} is accepting state.

Similarly DFA for REGEX_A (55*) is:

       |   5    |  epsilon*
   ----+--------+--------
    A  | B,C,E  |   A
    B  | C,D,E  |   B,C,E
    C  | C,D,E  |   C
    D  | C,D,E  |   C,D,E
    E  | C,D,E  |   C,E


            5(epsilon*)
   -------+---------------------
       A  |  B,C,E  
   B,C,E  |  C,D,E
   C,D,E  |  C,D,E

    {A} ---- 5 -----> {B,C,E}--5--->{C,D,E}
                                    {C,D,E}--5--->{C,D,E}
Note: {A} is start state and {C,D,E} is accepting state

Similarly DFA for REGEX_B (5*) is:

       |   5    |  epsilon*
   ----+--------+--------
    A  | B,C,D  |   A,B,D
    B  | B,C,D  |   B
    C  | B,C,D  |   B,C,D
    D  | B,C,D  |   B,D


            5(epsilon*)
   -------+---------------------
       A  |  B,C,D  
   B,C,D  |  B,C,D

    {A} ---- 5 -----> {B,C,D}
                      {B,C,D} --- 5 ---> {B,C,D}
Note: {A} is start state and {B,C,D} is accepting state

Conclusions:

DFA of REGX_A|REGX_B identical to DFA of REGX_A 
      -- implies REGEX_A is subset of REGEX_B
DFA of REGX_A|REGX_B is NOT identical to DFA of REGX_B 
      -- cannot infer about either gerexes.

In addition to antinome's answer:

Many of the constructs that are not part of the basic regex definition are still regular, and can be converted after parsing the regex (with a real parser, because the language of regex is not regular itself): (x?) to (x|), (x+) to (xx*), character classes like [a-d] to their corresponding union (a|b|c|d) etc. So one can use these constructs and still test whether one regex matches a subset of the other regex using the DFA comparison mentioned by antinome.

Some constructs, like back references, are not regular, and cannot be represented by NFA or DFA.

Even the seemingly simple problem of testing whether a regex with back references matches a particular string is NP-complete (http://perl.plover.com/NPC/NPC-3COL.html).

pip3 install https://github.com/leafstorm/lexington/archive/master.zip
python3
>>> from lexington.regex import Regex as R
>>> from lexington.regex import Null
>>> from functools import reduce
>>> from string import ascii_lowercase, digits
>>> a_z = reduce(lambda a, b: a | R(b), ascii_lowercase, Null)
>>> b_x = reduce(lambda a, b: a | R(b), ascii_lowercase[1:-2], Null)
>>> a_z | b_x == a_z
True
>>> m_n = R("m") | R("n")
>>> zero_nine = reduce(lambda a, b: a | R(b), digits, Null)
>>> m_n | zero_nine == m_n
False

Also tested successfully with Python 2. See also how to do it with a different library.

Alternatively, pip3 install https://github.com/ferno/greenery/archive/master.zip and:

from greenery.lego import parse as p
a_z = p("[a-z]")
b_x = p("[b-x]")
assert a_z | b_x == a_z
m_n = p("m|n")
zero_nine = p("[0-9]")
assert not m_n | zero_nine == m_n