'''
Created on 20-Jan-2022

@author: bthakur

Ref:
Integer Sequences: Divisibility, Lucas and Lehmer Sequences by Masum Billal, Samin Riasat
'''

import codecs
import inspect
import traceback
import random
import numpy as np
import pandas as pd
from sympy import (latex, Symbol, symbols,
                   Matrix,
                    Rational,
                    sin, cos, tan,
                    asin, acos, atan,
                    sinh, cosh, tanh,
                    log, exp, sqrt,
                   diff, factor, 
                   Function, Eq,
                   Add, 
                   simplify, solve, factorial,
                   UnevaluatedExpr as ueval
                   )

from sympy.core.numbers import One, Integer
from sympy.series import series

from xv.km.util import latex_fraction

from xv.util import strip_source_code_tabs, get_function_source_code


from xv.km.ui import SvgHelper
from xv.util.data import CacheUtil

import matplotlib.pyplot as plt
from xv.util import UtilClass, list_to_str
from xv.util import get_math_symb_desc, html_table

from ._set_algebra import SetAlgebra

class SetTheoryHelper(UtilClass):
    def __init__(self, **kwargs):
        super().__init__(**kwargs)

    def plot_venn_diagram(self):
        method_name = "_plot_venn_diagram"
        svg_str, plot_code = self._plot_specified_method(method_name)
        return svg_str, plot_code
    
    def _plot_venn_diagram(self):
        import matplotlib.pyplot as plt
        from matplotlib_venn import venn2

        rows = 1
        cols = 3
        fig = plt.figure(figsize=(6*cols, 6))
        gs = fig.add_gridspec(rows, cols)

        row_id = 0
        col_id = 0
        ax = fig.add_subplot(gs[row_id, col_id])
        v00 = venn2(ax = ax, subsets = (10, 2, 5), set_labels = ('Group A', 'Group B'))
        ax.set_title(f"joint sets")

        row_id = 0
        col_id = 1
        ax = fig.add_subplot(gs[row_id, col_id])
        v01 = venn2(ax = ax, subsets = (10, 7, 0), set_labels = ('Group A', 'Group B'))
        ax.set_title(f"disjoint sets")

        row_id = 0
        col_id = 2
        ax = fig.add_subplot(gs[row_id, col_id])
        v02 = venn2(ax = ax, subsets = (10, 0, 7), set_labels = ('Group A', 'Group B'))
        ax.set_title(f"subset")

        return fig, gs
    
    def _plot_specified_method(self, method_name):
        plt.ioff()
        
        if not hasattr(self, method_name):
            return None

        method = getattr(self, method_name)   
        plot_code = self._get_method_source_code(method)

        fig, ax = method()
        
        svgHelper = SvgHelper()
        svg_str = svgHelper.plt_to_svg()
        
        if not svg_str:
            if self.verbose:
                print(f"Error: svg_str could not be initialized")
            return None, plot_code
            
        plt.clf()
        plt.close()
        return svg_str, plot_code

    def _get_method_source_code(self, method, start = 1, end = -2):
        '''
        method to call. It must be a method, not name
        start: -1, line index to start
        end: -2, line index to end
        '''
        plot_code = get_function_source_code(method, 
                                             strip_tab = False)
        plot_code = plot_code[start : end]
        plot_code = strip_source_code_tabs(plot_code)
        
        plot_code.append("plt.show()")
        
        plot_code = "\n".join(plot_code)
        return plot_code
        pass
    
    
    def get_problem_set_properties(self):
        
        set_algebra = SetAlgebra(verbose = self.verbose)
        
        set_props = set_algebra.get_set_props()
        A = set_props.get('A', None)
        B = set_props.get('B', None)
        C = set_props.get('C', None)
        U = set_props.get('U', None)
        phi = set_props.get('phi', None)
        
    
        setA = set_props.get('setA', None)
        setB = set_props.get('setB', None)
        setC = set_props.get('setC', None)
        setNull = set()
        

        prob_s = f"""
        The sets are as follows:
        $\\displaystyle {latex(A)} = {latex(setA)}$
        $\\displaystyle {latex(B)} = {latex(setB)}$
        $\\displaystyle {latex(C)} = {latex(setC)}$

        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""
        The union of two sets is a set with elements of each set combined together.
        
        $$\\displaystyle 
        \\begin{{aligned}}
        
        {latex(A)} + {latex(B)} 
        
        & = {latex(A)} \\cup  {latex(B)} 
        \\\\ \\\\

        & = {latex(setA)} \\cup  {latex(setB)} 
        \\\\ \\\\
        
        & = {latex(setA.union(setB))} 
        
        \\end{{aligned}}
        $$

        """
        
        soln_stmts.append(soln_stmt.strip())


        soln_stmt = f"""
        The intersection (overlap) of two sets is a set with elements that are present in the both.
        
        $$\\displaystyle 
        \\begin{{aligned}}
        
        {latex(A)}{latex(B)} 
        
        & = {latex(A)} \\cap  {latex(B)} 
        \\\\ \\\\

        & = {latex(setA)} \\cap  {latex(setB)} 
        \\\\ \\\\
        
        & = {latex(setA.intersection(setB))} 
        
        \\end{{aligned}}
        $$

        """
        
        soln_stmts.append(soln_stmt.strip())
        
        
        soln_stmt = f"""
        The universal set is a set of all elements from all sets put together.
        As we have three sets here, the universal set
        
        $$\\displaystyle 
        \\begin{{aligned}}
        
        {latex(U)}
        
        & = {latex(A)} + {latex(B)} + {latex(C)} 
        \\\\ \\\\
        
        & = {latex(A)} \\cup  {latex(B)}  \\cup  {latex(C)} 
        \\\\ \\\\

        & = {latex(setA)} \\cup  {latex(setB)} \\cup  {latex(setC)} 
        \\\\ \\\\
        
        & = {latex(setA.union(setB).union(setC))} 
        
        \\end{{aligned}}
        $$

        """
        
        soln_stmts.append(soln_stmt.strip())
        
        
        soln_stmt = f"""
        A null set (empty set) is a set without elements.
        
        $$\\displaystyle 
        
        {latex(phi)} = {latex(set())} 
        $$

        """
        
        soln_stmts.append(soln_stmt.strip())
        
        
        answer_s = f"""
        """
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass
        

    
    def get_problem_identify_a_group(self):
        
        prob_s = f"""
        Prove theoretically and with an example that the set $\\mathbb{{Z}}
        $ (integers) equipped with addition is a group.
        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""

        
        1. Closure: For any two integers $a$ and $b$, their sum $a + b$ is also an integer.  
           $\\displaystyle 
           \\displaystyle \\forall a, b \\in \\mathbb{{Z}},\\ a + b \\in \\mathbb{{Z}}
            $  
           Examples:  
           $3 + (-5) = -2 \\in \\mathbb{{Z}}$  
           $7 + 4 = 11 \\in \\mathbb{{Z}}$  
           $-8 + (-6) = -14 \\in \\mathbb{{Z}}$
        
        2. Associativity: Addition of integers is associative.  
           $\\displaystyle 
           \\displaystyle \\forall a, b, c \\in \\mathbb{{Z}},\\ (a + b) + c = a + (b + c)
            $  
           Examples:  
           $(2 + 3) + 4 = 2 + (3 + 4) = 9$  
           $(-5 + 7) + (-3) = -5 + (7 + (-3)) = -1$  
           $(10 + (-2)) + 5 = 10 + ((-2) + 5) = 13$
        
        3. Identity Element: The integer $0$ is the identity element because for any integer $a$, $a + 0 = a$.  
           $\\displaystyle 
           \\displaystyle \\exists e = 0 \\in \\mathbb{{Z}}\\ \\text{{such that}}\\ \\forall a \\in \\mathbb{{Z}},\\ a + 0 = 0 + a = a
            $  
           Examples:  
           $7 + 0 = 7$  
           $-12 + 0 = -12$  
           $0 + 0 = 0$
        
        4. Inverse Element: For each integer $a$, there exists an inverse $-a$ such that $a + (-a) = 0$.  
           $\\displaystyle 
           \\displaystyle \\forall a \\in \\mathbb{{Z}},\\ \\exists (-a) \\in \\mathbb{{Z}}\\ \\text{{such that}}\\ a + (-a) = 0
            $  
           Examples:  
           $4 + (-4) = 0$  
           $-9 + 9 = 0$  
           $0 + 0 = 0$
        
        Since all four group axioms (closure, associativity, identity, and inverse) hold, we conclude that the set of integers $\\mathbb{{Z}}$ with addition forms a group.
        
        Thus, $\\mathbb{{Z}}$ under addition is a group.

                
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""
        We need to verify the group properties: closure, associativity, identity, 
        and existence of inverses for the set $\\mathbb{{Z}}$ under addition.
        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass
        
    
    def get_problem_identify_an_abelian_group(self):
        
        prob_s = f"""
        Prove theoretically and with examples that the set $\\mathbb{{Z}}
        $ (integers) equipped with addition is an abelian group.
        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""
        1. Closure: For any two integers $a$ and $b$, their sum $a + b$ is also an integer.  
           $\\displaystyle 
           \\displaystyle \\forall a, b \\in \\mathbb{{Z}},\\ a + b \\in \\mathbb{{Z}}
            $  
           Examples:  
           $2 + (-5) = -3 \\in \\mathbb{{Z}}$  
           $7 + 3 = 10 \\in \\mathbb{{Z}}$  
           $-8 + (-6) = -14 \\in \\mathbb{{Z}}$
        
        2. Associativity: Addition of integers is associative.  
           $\\displaystyle 
           \\displaystyle \\forall a, b, c \\in \\mathbb{{Z}},\\ (a + b) + c = a + (b + c)
            $  
           Examples:  
           $(1 + 2) + 4 = 1 + (2 + 4) = 7$  
           $(-3 + 6) + (-1) = -3 + (6 + (-1)) = 2$  
           $(10 + (-2)) + 5 = 10 + ((-2) + 5) = 13$
        
        3. Identity Element: The integer $0$ is the identity element because for any integer $a$, $a + 0 = a$.  
           $\\displaystyle 
           \\displaystyle \\exists e = 0 \\in \\mathbb{{Z}}\\ \\text{{such that}}\\ \\forall a \\in \\mathbb{{Z}},\\ a + 0 = 0 + a = a
            $  
           Examples:  
           $5 + 0 = 5$  
           $-9 + 0 = -9$  
           $0 + 0 = 0$
        
        4. Inverse Element: For each integer $a$, there exists an inverse $-a$ such that $a + (-a) = 0$.  
           $\\displaystyle 
           \\displaystyle \\forall a \\in \\mathbb{{Z}},\\ \\exists (-a) \\in \\mathbb{{Z}}\\ \\text{{such that}}\\ a + (-a) = 0
            $  
           Examples:  
           $3 + (-3) = 0$  
           $-5 + 5 = 0$  
           $0 + 0 = 0$
        
        5. Commutativity: Addition of integers is commutative, meaning $a + b = b + a$.  
           $\\displaystyle 
           \\displaystyle \\forall a, b \\in \\mathbb{{Z}},\\ a + b = b + a
            $  
           Examples:  
           $4 + 7 = 7 + 4 = 11$  
           $-3 + 2 = 2 + (-3) = -1$  
           $6 + (-6) = (-6) + 6 = 0$
        
        Since all five properties (closure, associativity, identity, inverse, and commutativity) hold, we conclude that the set of integers $\\mathbb{{Z}}$ with addition forms an abelian group.
        
        Thus, $\\mathbb{{Z}}$ under addition is an abelian group.
        
                
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""
        To prove that $\\mathbb{{Z}}$ under addition is an abelian group, 
        we need to verify the group properties 
        (closure, associativity, identity, and inverse) as well as the commutative property.
        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass
        
    def get_problem_lagrange_s_theorem_of_subgroup(self):
        
        prob_s = f"""
        Prove by example that if $H$ is a subgroup of $G
        $, then $|H|$ divides $|G|$ (Lagrange's Theorem).
        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""
        Let us take an example to demonstrate this theorem.
        
        Consider the group $G = \\mathbb{{Z}}_6$, the set of integers under addition modulo 6. The group $\\mathbb{{Z}}_6 = \\{{0, 1, 2, 3, 4, 5\\}}$ has 6 elements, so $|G| = 6$.
        
        Now, let's consider the subgroup $H = \\{{0, 3\\}}$ of $\\mathbb{{Z}}_6$. The set $H$ contains 2 elements, so $|H| = 2$.
        
        By Lagrange's Theorem, the order of $H$ should divide the order of $G$. We can check that:
        $\\displaystyle 
        \\frac{{|G|}}{{|H|}} = \\frac{{6}}{{2}} = 3
         $
        Since $2$ divides $6$, this satisfies Lagrange's Theorem.
        
        Examples:
        
        1. $G = \\mathbb{{Z}}_6 = \\{{0, 1, 2, 3, 4, 5\\}}$, and $H = \\{{0, 3\\}}$ is a subgroup with $|H| = 2$. Here, $2$ divides $6$.
        2. Consider another example where $G = \\mathbb{{Z}}_4 = \\{{0, 1, 2, 3\\}}$ and $H = \\{{0, 2\\}}$ is a subgroup with $|H| = 2$. Again, $2$ divides $4$.
        3. If $G = \\mathbb{{Z}}_8 = \\{{0, 1, 2, 3, 4, 5, 6, 7\\}}$ and $H = \\{{0, 4\\}}$ is a subgroup with $|H| = 2$, then $2$ divides $8$.
        
        Thus, in each example, the order of the subgroup divides the order of the group, verifying Lagrange’s Theorem.
        
                
                
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""
        Lagrange’s Theorem states that if $H$ is a subgroup of a finite group $G$, 
        then the order (size) of $H$ divides the order of $G$.
        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass
        
    def get_problem_prove_example_subgroup_and_lagrange_s_theorem(self):
        
        prob_s = f"""
        Create a subgroup of a finite group of integers under addition, 
        and prove why they are subgroups. Also prove that they follow Lagrange's Theorem.
        """
        prob_s = prob_s.replace("\n", " ").strip()
        
        
        soln_stmts = []
        
        soln_stmt = f"""
        <b>Group:</b> 
        
        Let the operation be $\\text {{modulo 6 under addition}} $, i.e., 
        $ z = (x + y) \% 6 $
        where % represents modulo or remainder after dividing by $6$.
        
        Let the group be $\\mathbb{{Z}}_6 = \\{{0, 1, 2, 3, 4, 5\\}}$.
        
        <b>Subgroup:</b> 
        
        $H_1 = \\{{0, 2, 4\\}}$ in $\\mathbb{{Z}}_6$
        
        This is a finite subgroup of $\\mathbb{{Z}}_6$ (integers modulo 6 under addition). 
        
        <b>Verification of Subgroup Properties:</b> 
 
        1. Closure: The sum of any two elements in $H_1$ is still in $H_1$.  
           Examples: 
           $(2 + 2) \% 6 = 4 \% 6 = 4 \\in H_1 $
           $(4 + 2) \% 6 = 6 \% 6 = 0 \\in H_1 $
           $(0 + 2) \% 6 = 4 \% 6 = 4 \\in H_1 $
        
        2. Identity: The identity element is $0$, which is in $H_1$.
           Examples:  
           $(0 + 0) \% 6 = 0 \% 6 = 0 \\in H_1 $
           $(0 + 2) \% 6 = 2 \% 6 = 2 \\in H_1 $
           $(0 + 4) \% 6 = 4 \% 6 = 4 \\in H_1 $
        
        3. Inverses: Each element has an inverse within the set.  
           Examples:  
           $(0 + 0) \% 6 = 0 \% 6 = 0 \\text{{ identity element in }} H_1 $
           $(2 + 4) \% 6 = 6 \% 6 = 0 \\text{{ identity element in }} H_1 $
           $(4 + 2) \% 6 = 6 \% 6 = 0 \\text{{ identity element in }} H_1 $
        
        Thus, $H_1 = \\{{0, 2, 4\\}}$ is a valid subgroup of $\\mathbb{{Z}}_6$.
        
        <b>Applying Lagrange’s Theorem:</b> 
        
        Order of a group is number of elements in it.
          
        Order of the group $\\mathbb{{Z}}_6 = | \\mathbb{{Z}}_6 | = 6$
        
        Order of the subgroup $H_1 = |H_1| = 3$
         
        According to Lagrange’s Theorem, $|H_1|$ divides $| \\mathbb{{Z}}_6 |$.  
        
        $\\displaystyle 
        \\frac{{ | \\mathbb{{Z}}_6 | }}{{|H_1|}} = \\frac{{6}}{{3}} = 2
         $
        
        Thus, $H_1$ satisfies Lagrange’s Theorem.
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""

        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass



        
    def get_problem_identify_a_ring(self):
        
        prob_s = f"""
        Prove theoretically and with examples that the set $\\mathbb{{Q}}
        $ (rational numbers) equipped with addition is a ring.
        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""
        <b>Definition of a Ring:</b>
        A set $S$ equipped with two operations $+$ (addition) and $\\cdot$ (multiplication) is a ring if it satisfies the following properties:
        
        1. Abelian Additive Group: $S$ is an abelian group under $+$, with identity element $0$.
        
        2. Associativity of Multiplication: For any $x, y, z \\in S$,  
           $\\displaystyle 
           x \\cdot (y \\cdot z) = (x \\cdot y) \\cdot z
            $
        
        3. Identity of Multiplication: There exists an element $1 \\in S$ such that for any $x \\in S$,  
           $\\displaystyle 
           x \\cdot 1 = 1 \\cdot x = x
            $
        
        4. Distributivity of Multiplication over Addition: For any $x, y, z \\in S$,  
           $\\displaystyle 
           x \\cdot (y + z) = x \\cdot y + x \\cdot z
           \\quad \\text{{and}} \\quad
           (y + z) \\cdot x = y \\cdot x + z \\cdot x
            $
        
        <b>Step-by-Step Proof for $\\mathbb{{Q}}$ as a Ring:</b>
        
        <b>1. Abelian Additive Group:</b>
        We have already shown that $\\mathbb{{Q}}$, the set of rational numbers, forms an abelian group under addition. Let’s summarize the group properties:
        
        &#9679; Closure: The sum of any two rational numbers is a rational number.
          $\\displaystyle 
          \\forall a, b \\in \\mathbb{{Q}},\\ a + b \\in \\mathbb{{Q}}
           $
          Example: $\\frac{{1}}{{3}} + \\frac{{2}}{{5}} = \\frac{{11}}{{15}} \\in \\mathbb{{Q}}$.
        
        &#9679; Associativity: Addition is associative.
          $\\displaystyle 
          \\forall a, b, c \\in \\mathbb{{Q}},\\ (a + b) + c = a + (b + c)
           $
          Example:  
          $\\left(\\frac{{1}}{{4}} + \\frac{{1}}{{2}}\\right) + \\frac{{1}}{{3}} = \\frac{{13}}{{12}}$  
          and  
          $\\frac{{1}}{{4}} + \\left(\\frac{{1}}{{2}} + \\frac{{1}}{{3}}\\right) = \\frac{{13}}{{12}}$.
        
        &#9679; Identity Element: The additive identity is $0$.
          $\\displaystyle 
          \\exists 0 \\in \\mathbb{{Q}}\\ \\text{{such that}}\\ \\forall a \\in \\mathbb{{Q}},\\ a + 0 = a
           $
          Example:  
          $\\frac{{3}}{{7}} + 0 = \\frac{{3}}{{7}}$.
        
        &#9679; Inverses: For every rational number $a$, there exists an additive inverse $-a$ such that $a + (-a) = 0$.
          $\\displaystyle 
          \\forall a \\in \\mathbb{{Q}},\\ \\exists (-a) \\in \\mathbb{{Q}}\\ \\text{{such that}}\\ a + (-a) = 0
           $
          Example:  
          The inverse of $\\frac{{2}}{{5}}$ is $-\\frac{{2}}{{5}}$, and  
          $\\frac{{2}}{{5}} + \\left(-\\frac{{2}}{{5}}\\right) = 0$.
        
        &#9679; Commutativity: Addition of rational numbers is commutative.
          $\\displaystyle 
          \\forall a, b \\in \\mathbb{{Q}},\\ a + b = b + a
           $
          Example:  
          $\\frac{{1}}{{2}} + \\frac{{1}}{{3}} = \\frac{{5}}{{6}} = \\frac{{1}}{{3}} + \\frac{{1}}{{2}}$.
        
        Thus, $\\mathbb{{Q}}$ forms an abelian group under addition.
        
        <b>2. Associativity of Multiplication:</b>
        Multiplication in $\\mathbb{{Q}}$ is associative. For any $x, y, z \\in \\mathbb{{Q}}$,
        $\\displaystyle 
        x \\cdot (y \\cdot z) = (x \\cdot y) \\cdot z
         $
        Example:  
        $\\displaystyle 
        \\left(\\frac{{1}}{{2}} \\times \\frac{{3}}{{4}}\\right) \\times \\frac{{5}}{{6}} = \\frac{{3}}{{8}} \\times \\frac{{5}}{{6}} = \\frac{{15}}{{48}}
         $
        and  
        $\\displaystyle 
        \\frac{{1}}{{2}} \\times \\left(\\frac{{3}}{{4}} \\times \\frac{{5}}{{6}}\\right) = \\frac{{1}}{{2}} \\times \\frac{{15}}{{24}} = \\frac{{15}}{{48}}
         $
        
        Thus, multiplication is associative in $\\mathbb{{Q}}$.
        
        <b>3. Identity of Multiplication:</b>
        The multiplicative identity in $\\mathbb{{Q}}$ is $1$. For any $x \\in \\mathbb{{Q}}$,
        $\\displaystyle 
        x \\cdot 1 = 1 \\cdot x = x
         $
        Example:  
        $\\displaystyle 
        \\frac{{3}}{{5}} \\times 1 = 1 \\times \\frac{{3}}{{5}} = \\frac{{3}}{{5}}
         $
        
        <b>4. Distributivity of Multiplication over Addition:</b>
        Multiplication distributes over addition in $\\mathbb{{Q}}$. For any $x, y, z \\in \\mathbb{{Q}}$,
        
        &#9679; Left Distributive:
          $\\displaystyle 
          x \\cdot (y + z) = x \\cdot y + x \\cdot z
           $
          Example:  
          $\\displaystyle 
          \\frac{{1}}{{2}} \\times \\left(\\frac{{1}}{{3}} + \\frac{{1}}{{4}}\\right) = \\frac{{1}}{{2}} \\times \\frac{{7}}{{12}} = \\frac{{7}}{{24}}
           $
          and  
          $\\displaystyle 
          \\left(\\frac{{1}}{{2}} \\times \\frac{{1}}{{3}}\\right) + \\left(\\frac{{1}}{{2}} \\times \\frac{{1}}{{4}}\\right) = \\frac{{1}}{{6}} + \\frac{{1}}{{8}} = \\frac{{7}}{{24}}
           $
        
        &#9679; Right Distributive:
          $\\displaystyle 
          (y + z) \\cdot x = y \\cdot x + z \\cdot x
           $
          Example:  
          $\\displaystyle 
          \\left(\\frac{{1}}{{3}} + \\frac{{1}}{{4}}\\right) \\times \\frac{{1}}{{2}} = \\frac{{7}}{{12}} \\times \\frac{{1}}{{2}} = \\frac{{7}}{{24}}
           $
          and  
          $\\displaystyle 
          \\left(\\frac{{1}}{{3}} \\times \\frac{{1}}{{2}}\\right) + \\left(\\frac{{1}}{{4}} \\times \\frac{{1}}{{2}}\\right) = \\frac{{1}}{{6}} + \\frac{{1}}{{8}} = \\frac{{7}}{{24}}
           $
        
        Thus, multiplication distributes over addition in $\\mathbb{{Q}}$.
        
        <b>Conclusion:</b>
        Since $\\mathbb{{Q}}$ forms an abelian group under addition, multiplication is associative, there is an identity for multiplication, and multiplication distributes over addition, the set of rational numbers $\\mathbb{{Q}}$ equipped with addition and multiplication is a ring.
        
                        
                        
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""
        NA
        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass


    def get_problem_identify_a_field(self):
        
        prob_s = f"""
        Prove theoretically and with examples that the set $\\mathbb{{Q}}
        $ (rational numbers) equipped with addition and multiplication is a field.
        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""
    
        
        
        <b>Definition of a Field:</b>
        
        A set $S$ equipped with two operations, $+$ (addition) and $\\cdot$ (multiplication), is called a field if it satisfies the following properties:
        
        <b>It is a Ring:</b>
        1. Abelian Additive Group: $S$ is an abelian group under $+$, with identity element $0$.
        
        2. Associativity of Multiplication: For any $x, y, z \\in S$,  
           $\\displaystyle 
           x \\cdot (y \\cdot z) = (x \\cdot y) \\cdot z
            $
        
        3. Identity of Multiplication: There exists an element $1 \\in S$ such that for any $x \\in S$,  
           $\\displaystyle 
           x \\cdot 1 = 1 \\cdot x = x
            $
        
        4. Distributivity of Multiplication over Addition: For any $x, y, z \\in S$,  
           $\\displaystyle 
           x \\cdot (y + z) = x \\cdot y + x \\cdot z
           \\quad \\text{{and}} \\quad
           (y + z) \\cdot x = y \\cdot x + z \\cdot x
            $
        
        <b>Additional Properties of a Field:</b>
        5. Commutativity of Multiplication: For any $x, y \\in S$,  
           $\\displaystyle 
           x \\cdot y = y \\cdot x
            $
        
        6. Inverses for Multiplication: For any $x \\in S \\setminus \\{{0\\}}$, there exists an element $x^{{-1}} \\in S$ such that  
           $\\displaystyle 
           x \\cdot x^{{-1}} = x^{{-1}} \\cdot x = 1
            $
        
        <b>Proof that $\\mathbb{{Q}}$ is a Field:</b>
        
        We will now verify that $\\mathbb{{Q}}$ (the set of rational numbers) satisfies all the properties of a field.
        
        <b>1. Abelian Additive Group:</b>
        We already know that $\\mathbb{{Q}}$ forms an abelian group under addition, with $0$ as the additive identity. This was proved earlier, where we showed that:
        &#9679; Closure: The sum of any two rational numbers is a rational number.
        &#9679; Associativity: Addition is associative.
        &#9679; Identity Element: The additive identity is $0$.
        &#9679; Inverses: Each element has an additive inverse.
        &#9679; Commutativity: Addition is commutative.
        
        Thus, $\\mathbb{{Q}}$ forms an abelian group under addition.
        
        <b>2. Associativity of Multiplication:</b>
        Multiplication in $\\mathbb{{Q}}$ is associative. For any $x, y, z \\in \\mathbb{{Q}}$,  
        $\\displaystyle 
        x \\cdot (y \\cdot z) = (x \\cdot y) \\cdot z
         $
        Example:  
        $\\displaystyle 
        \\left(\\frac{{1}}{{2}} \\times \\frac{{3}}{{4}}\\right) \\times \\frac{{5}}{{6}} = \\frac{{3}}{{8}} \\times \\frac{{5}}{{6}} = \\frac{{15}}{{48}}
         $
        and  
        $\\displaystyle 
        \\frac{{1}}{{2}} \\times \\left(\\frac{{3}}{{4}} \\times \\frac{{5}}{{6}}\\right) = \\frac{{1}}{{2}} \\times \\frac{{15}}{{24}} = \\frac{{15}}{{48}}
         $
        
        <b>3. Identity of Multiplication:</b>
        The multiplicative identity in $\\mathbb{{Q}}$ is $1$. For any $x \\in \\mathbb{{Q}}$,  
        $\\displaystyle 
        x \\cdot 1 = 1 \\cdot x = x
         $
        Example:  
        $\\displaystyle 
        \\frac{{3}}{{5}} \\times 1 = 1 \\times \\frac{{3}}{{5}} = \\frac{{3}}{{5}}
         $
        
        <b>4. Distributivity of Multiplication over Addition:</b>
        Multiplication distributes over addition in $\\mathbb{{Q}}$. For any $x, y, z \\in \\mathbb{{Q}}$,  
        $\\displaystyle 
        x \\cdot (y + z) = x \\cdot y + x \\cdot z
         $
        Example:  
        $\\displaystyle 
        \\frac{{1}}{{2}} \\times \\left(\\frac{{1}}{{3}} + \\frac{{1}}{{4}}\\right) = \\frac{{1}}{{2}} \\times \\frac{{7}}{{12}} = \\frac{{7}}{{24}}
         $
        and  
        $\\displaystyle 
        \\frac{{1}}{{2}} \\times \\frac{{1}}{{3}} + \\frac{{1}}{{2}} \\times \\frac{{1}}{{4}} = \\frac{{1}}{{6}} + \\frac{{1}}{{8}} = \\frac{{7}}{{24}}
         $
        
        <b>5. Commutativity of Multiplication:</b>
        Multiplication in $\\mathbb{{Q}}$ is commutative. For any $x, y \\in \\mathbb{{Q}}$,  
        $\\displaystyle 
        x \\cdot y = y \\cdot x
         $
        Example:  
        $\\displaystyle 
        \\frac{{3}}{{5}} \\times \\frac{{2}}{{7}} = \\frac{{6}}{{35}}
         $
        and  
        $\\displaystyle 
        \\frac{{2}}{{7}} \\times \\frac{{3}}{{5}} = \\frac{{6}}{{35}}
         $
        
        <b>6. Inverses for Multiplication:</b>
        For every non-zero rational number $x \\in \\mathbb{{Q}}$, there exists a multiplicative inverse $x^{{-1}} = \\frac{{1}}{{x}}$ such that  
        $\\displaystyle 
        x \\cdot x^{{-1}} = 1
         $
        Example:  
        The inverse of $\\frac{{3}}{{5}}$ is $\\frac{{5}}{{3}}$, and  
        $\\displaystyle 
        \\frac{{3}}{{5}} \\times \\frac{{5}}{{3}} = 1
         $
        This is true for any non-zero rational number.
        
        <b>Conclusion:</b>
        Since $\\mathbb{{Q}}$ satisfies all the properties of a ring and the additional properties required for a field (commutativity of multiplication and the existence of multiplicative inverses), we conclude that $\\mathbb{{Q}}$ is a field under addition and multiplication.
        
        
                
                        
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""
        NA
        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass

    def get_problem_identify_a_vector_space(self):
        
        prob_s = f"""
        Prove theoretically and with examples that $
        \\mathbb{{R}}^n$ is a vector space over the field $\\mathbb{{R}}$.
        """
        
        
        soln_stmts = []
        
        soln_stmt = f"""
        <b>Definition of a Vector Space:</b>
        
        A set $V$ is called a vector space over a field $F$ if it satisfies the following properties:
        
        1. Abelian Additive Group: $V$ is an abelian group under vector addition $+$, with identity element $\\mathbf{{0}}$.
        
        2. Closure under Scalar Multiplication: For any scalar $a \\in F$ and any vector $\\mathbf{{v}} \\in V$, the scalar multiple $a\\mathbf{{v}}$ is in $V$.
        
        3. Properties of Scalar Multiplication: For all scalars $a, b \\in F$ and vectors $\\mathbf{{u}}, \\mathbf{{v}} \\in V$,
           &#9679; (i) Distributivity of Scalar over Vector Addition:
             $\\displaystyle 
             a(\\mathbf{{u}} + \\mathbf{{v}}) = a\\mathbf{{u}} + a\\mathbf{{v}}
              $
           &#9679; (ii) Distributivity of Scalars over Scalar Addition:
             $\\displaystyle 
             (a + b)\\mathbf{{v}} = a\\mathbf{{v}} + b\\mathbf{{v}}
              $
           &#9679; (iii) Associativity of Scalar Multiplication:
             $\\displaystyle 
             a(b\\mathbf{{v}}) = (ab)\\mathbf{{v}}
              $
           &#9679; (iv) Multiplication by Zero Scalar:
             $\\displaystyle 
             0\\mathbf{{v}} = \\mathbf{{0}}
              $
           &#9679; (v) Multiplication by Unity Scalar:
             $\\displaystyle 
             1\\mathbf{{v}} = \\mathbf{{v}}
              $
        
        
        
        <b>Step-by-Step Proof for $\\displaystyle  V = \\mathbb{{R}}^n $ as a Vector Space over $\\displaystyle  F = \\mathbb{{R}} $:</b>
        
        We will demonstrate that the set of all $n$-tuples of real numbers, $\\mathbb{{R}}^n$, is a vector space over the field of real numbers $\\mathbb{{R}}$.
        
        <b>1. Abelian Additive Group:</b>
        
        We need to show that $\\mathbb{{R}}^n$ forms an abelian group under vector addition.
        
        &#9679; Closure: The sum of any two vectors in $\\mathbb{{R}}^n$ is also in $\\mathbb{{R}}^n$.
          $\\displaystyle 
          \\forall \\mathbf{{u}}, \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ \\mathbf{{u}} + \\mathbf{{v}} \\in \\mathbb{{R}}^n
           $
          Example:
          $\\displaystyle 
          \\mathbf{{u}} = (1, 2, 3),\\ \\mathbf{{v}} = (4, 5, 6) \\implies \\mathbf{{u}} + \\mathbf{{v}} = (5, 7, 9) \\in \\mathbb{{R}}^3
           $
        
        &#9679; Associativity: Vector addition is associative.
          $\\displaystyle 
          \\forall \\mathbf{{u}}, \\mathbf{{v}}, \\mathbf{{w}} \\in \\mathbb{{R}}^n,\\ (\\mathbf{{u}} + \\mathbf{{v}}) + \\mathbf{{w}} = \\mathbf{{u}} + (\\mathbf{{v}} + \\mathbf{{w}})
           $
          Example:
          $\\displaystyle 
          \\begin{{align*}}
          \\mathbf{{u}} &= (1, 2),\\ \\mathbf{{v}} = (3, 4),\\ \\mathbf{{w}} = (5, 6) \\\\
          (\\mathbf{{u}} + \\mathbf{{v}}) + \\mathbf{{w}} &= (1+3, 2+4) + (5, 6) = (4, 6) + (5, 6) = (9, 12) \\\\
          \\mathbf{{u}} + (\\mathbf{{v}} + \\mathbf{{w}}) &= (1, 2) + (3+5, 4+6) = (1, 2) + (8, 10) = (9, 12)
          \\end{{align*}}
           $
        
        &#9679; Identity Element: The zero vector $\\mathbf{{0}} = (0, 0, \\dots, 0)$ is the additive identity.
          $\\displaystyle 
          \\exists \\mathbf{{0}} \\in \\mathbb{{R}}^n\\ \\text{{such that}}\\ \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ \\mathbf{{v}} + \\mathbf{{0}} = \\mathbf{{v}}
           $
          Example:
          $\\displaystyle 
          \\mathbf{{v}} = (3, 7, 2) \\implies \\mathbf{{v}} + \\mathbf{{0}} = (3+0, 7+0, 2+0) = (3, 7, 2)
           $
        
        &#9679; Inverses: Every vector has an additive inverse.
          $\\displaystyle 
          \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ \\exists -\\mathbf{{v}} \\in \\mathbb{{R}}^n\\ \\text{{such that}}\\ \\mathbf{{v}} + (-\\mathbf{{v}}) = \\mathbf{{0}}
           $
          Example:
          $\\displaystyle 
          \\mathbf{{v}} = (3, -5, 2) \\implies -\\mathbf{{v}} = (-3, 5, -2),\\ \\mathbf{{v}} + (-\\mathbf{{v}}) = (0, 0, 0)
           $
        
        &#9679; Commutativity: Vector addition is commutative.
          $\\displaystyle 
          \\forall \\mathbf{{u}}, \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ \\mathbf{{u}} + \\mathbf{{v}} = \\mathbf{{v}} + \\mathbf{{u}}
           $
          Example:
          $\\displaystyle 
          \\mathbf{{u}} = (1, 2),\\ \\mathbf{{v}} = (3, 4) \\implies \\mathbf{{u}} + \\mathbf{{v}} = (4, 6) = \\mathbf{{v}} + \\mathbf{{u}}
           $
        
        Thus, $\\mathbb{{R}}^n$ forms an abelian group under vector addition.
        
        <b>2. Closure under Scalar Multiplication:</b>
        
        For any scalar $a \\in \\mathbb{{R}}$ and vector $\\mathbf{{v}} \\in \\mathbb{{R}}^n$, the product $a\\mathbf{{v}}$ is in $\\mathbb{{R}}^n$.
        $\\displaystyle 
        \\forall a \\in \\mathbb{{R}},\\ \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ a\\mathbf{{v}} \\in \\mathbb{{R}}^n
         $
        Proof: If $\\mathbf{{v}} = (v_1, v_2, \\dots, v_n)$, then $a\\mathbf{{v}} = (a v_1, a v_2, \\dots, a v_n) \\in \\mathbb{{R}}^n$.
        
        Example:
        $\\displaystyle 
        a = 2,\\ \\mathbf{{v}} = (1, -3, 4) \\implies a\\mathbf{{v}} = (2, -6, 8) \\in \\mathbb{{R}}^3
         $
        
        <b>3. Properties of Scalar Multiplication:</b>
        
        We will verify each property individually.
        
        <b>(i) Distributivity of Scalar over Vector Addition:</b>
        $\\displaystyle 
        \\forall a \\in \\mathbb{{R}},\\ \\forall \\mathbf{{u}}, \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ a(\\mathbf{{u}} + \\mathbf{{v}}) = a\\mathbf{{u}} + a\\mathbf{{v}}
         $
        Proof:
        $\\displaystyle 
        \\begin{{align*}}
        a(\\mathbf{{u}} + \\mathbf{{v}}) &= a(u_1 + v_1, u_2 + v_2, \\dots, u_n + v_n) \\\\
        &= (a(u_1 + v_1), a(u_2 + v_2), \\dots, a(u_n + v_n)) \\\\
        &= (a u_1 + a v_1, a u_2 + a v_2, \\dots, a u_n + a v_n) \\\\
        &= (a u_1, a u_2, \\dots, a u_n) + (a v_1, a v_2, \\dots, a v_n) \\\\
        &= a\\mathbf{{u}} + a\\mathbf{{v}}
        \\end{{align*}}
         $
        Example:
        $\\displaystyle 
        \\begin{{align*}}
        a &= 3,\\ \\mathbf{{u}} = (1, 2),\\ \\mathbf{{v}} = (4, 5) \\\\
        a(\\mathbf{{u}} + \\mathbf{{v}}) &= 3(1+4, 2+5) = 3(5, 7) = (15, 21) \\\\
        a\\mathbf{{u}} + a\\mathbf{{v}} &= (3 \\cdot 1, 3 \\cdot 2) + (3 \\cdot 4, 3 \\cdot 5) = (3, 6) + (12, 15) = (15, 21)
        \\end{{align*}}
         $
        
        <b>(ii) Distributivity of Scalars over Scalar Addition:</b>
        $\\displaystyle 
        \\forall a, b \\in \\mathbb{{R}},\\ \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ (a + b)\\mathbf{{v}} = a\\mathbf{{v}} + b\\mathbf{{v}}
         $
        Proof:
        $\\displaystyle 
        \\begin{{align*}}
        (a + b)\\mathbf{{v}} &= ((a + b)v_1, (a + b)v_2, \\dots, (a + b)v_n) \\\\
        &= (a v_1 + b v_1, a v_2 + b v_2, \\dots, a v_n + b v_n) \\\\
        &= (a v_1, a v_2, \\dots, a v_n) + (b v_1, b v_2, \\dots, b v_n) \\\\
        &= a\\mathbf{{v}} + b\\mathbf{{v}}
        \\end{{align*}}
         $
        Example:
        $\\displaystyle 
        \\begin{{align*}}
        a &= 2,\\ b = 3,\\ \\mathbf{{v}} = (1, -1) \\\\
        (a + b)\\mathbf{{v}} &= (5)(1, -1) = (5, -5) \\\\
        a\\mathbf{{v}} + b\\mathbf{{v}} &= (2 \\cdot 1, 2 \\cdot -1) + (3 \\cdot 1, 3 \\cdot -1) = (2, -2) + (3, -3) = (5, -5)
        \\end{{align*}}
         $
        
        <b>(iii) Associativity of Scalar Multiplication:</b>
        $\\displaystyle 
        \\forall a, b \\in \\mathbb{{R}},\\ \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ a(b\\mathbf{{v}}) = (a b)\\mathbf{{v}}
         $
        Proof:
        $\\displaystyle 
        \\begin{{align*}}
        a(b\\mathbf{{v}}) &= a(b v_1, b v_2, \\dots, b v_n) \\\\
        &= (a b v_1, a b v_2, \\dots, a b v_n) \\\\
        &= (a b)\\mathbf{{v}}
        \\end{{align*}}
         $
        Example:
        $\\displaystyle 
        \\begin{{align*}}
        a &= 4,\\ b = 2,\\ \\mathbf{{v}} = (3, 5) \\\\
        a(b\\mathbf{{v}}) &= 4(2 \\cdot 3, 2 \\cdot 5) = 4(6, 10) = (24, 40) \\\\
        (a b)\\mathbf{{v}} &= (8)(3, 5) = (24, 40)
        \\end{{align*}}
         $
        
        <b>(iv) Multiplication by Zero Scalar:</b>
        $\\displaystyle 
        \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ 0\\mathbf{{v}} = \\mathbf{{0}}
         $
        Proof:
        $\\displaystyle 
        0\\mathbf{{v}} = (0 \\cdot v_1, 0 \\cdot v_2, \\dots, 0 \\cdot v_n) = (0, 0, \\dots, 0) = \\mathbf{{0}}
         $
        Example:
        $\\displaystyle 
        \\mathbf{{v}} = (7, -3, 5) \\implies 0\\mathbf{{v}} = (0, 0, 0) = \\mathbf{{0}}
         $
        
        <b>(v) Multiplication by Unity Scalar:</b>
        $\\displaystyle 
        \\forall \\mathbf{{v}} \\in \\mathbb{{R}}^n,\\ 1\\mathbf{{v}} = \\mathbf{{v}}
         $
        Proof:
        $\\displaystyle 
        1\\mathbf{{v}} = (1 \\cdot v_1, 1 \\cdot v_2, \\dots, 1 \\cdot v_n) = (v_1, v_2, \\dots, v_n) = \\mathbf{{v}}
         $
        Example:
        $\\displaystyle 
        \\mathbf{{v}} = (4, -2) \\implies 1\\mathbf{{v}} = (4, -2) = \\mathbf{{v}}
         $
        
        <b>Conclusion:</b>
        
        Since $\\mathbb{{R}}^n$ satisfies all the properties of an abelian group under vector addition, scalar multiplication is closed and adheres to the necessary properties, we conclude that $\\mathbb{{R}}^n$ is a vector space over the field $\\mathbb{{R}}$.

                        
        """
        
        soln_stmts.append(soln_stmt.strip())

        math_symb_desc_html_table = get_math_symb_desc(soln_stmts, as_html_table = True)

        soln_stmt = f"""
        {math_symb_desc_html_table}
        """
        
        soln_stmts.insert(0, soln_stmt.strip())
        
        
        
        answer_s = f"""

        """
        answer_s = answer_s.replace("\n", " ").strip()
        
        soln_s = "\n\n".join(soln_stmts)
        
        result = {
            'prob_s': prob_s.strip(),
            'soln_s': soln_s.strip(),
            'answer_s': answer_s.strip(),
            }
        
        return result
        pass


    pass 
#class SetTheoryHelper:
