Sympy：手动处理等式

我目前正在学习数学课程，我的目标是理解概念和过程，而不是尽快解决问题集。在解方程时，我希望能够自己研究它们，而不是让它们帮我解。

假设我们有一个非常简单的方程z + 1 = 4- 如果我自己解决这个问题，我显然会从两边减去1，但我不知道是否sympy提供了一种简单的方法来做到这一点。目前我能想到的最好的解决方案是：

from sympy import *
z = symbols('z')
eq1 = Eq(z + 1, 4)
Eq(eq1.lhs - 1, eq1.rhs - 1)
# Output:
# z == 3

哪里表达比较明显eq1 - 1只从左侧减去。我怎样才能使用 sympy 来像这样一步一步地解决等式（即没有得到solve()刚刚给我答案的方法）？任何指向 sympy 等式实际上可能进行的操作的指针都将受到赞赏。

有一个“do”方法和讨论https://github.com/sympy/sympy/issues/5031#issuecomment-36996878这将允许您对等式的两边进行“操作”。它尚未被接受为 SymPy 的补充，但它是一个可以使用的简单附加组件。为了方便起见，将其粘贴在这里：

def do(self, e, i=None, doit=False):
    """Return a new Eq using function given or a model
    model expression in which a variable represents each
    side of the expression.

    Examples
    ========

    >>> from sympy import Eq
    >>> from sympy.abc import i, x, y, z
    >>> eq = Eq(x, y)

    When the argument passed is an expression with one
    free symbol that symbol is used to indicate a "side"
    in the Eq and an Eq will be returned with the sides
    from self replaced in that expression. For example, to
    add 2 to both sides:

    >>> eq.do(i + 2)
    Eq(x + 2, y + 2)

    To add x to both sides:

    >>> eq.do(i + x)
    Eq(2*x, x + y)

    In the preceding it was actually ambiguous whether x or i
    was to be added but the rule is that any symbol that are
    already in the expression are not to be interpreted as the
    dummy variable. If we try to add z to each side, however, an 
    error is raised because now it is unclear whether i or z is being
    added:

    >>> eq.do(i + z)
    Traceback (most recent call last):
    ...
    ValueError: not sure what symbol is being used to represent a side

    The ambiguity must be resolved by indicating with another parameter 
    which is the dummy variable representing a side:

    >>> eq.do(i + z, i)
    Eq(x + z, y + z)

    Alternatively, if only one Dummy symbol appears in the expression then
    it will be automatically used to represent a side of the Eq.

    >>> eq.do(2*Dummy() + z)
    Eq(2*x + z, 2*y + z)

    Operations like differentiation must be passed as a
    lambda:

    >>> Eq(x, y).do(lambda i: i.diff(x))
    Eq(1, 0)

    Because doit=False by default, the result is not evaluated. to
    evaluate it, either use the doit method or pass doit=True.

    >>> _.doit == Eq(x, y).do(lambda i: i.diff(x), doit=True)
    True
    """
    if not isinstance(e, (FunctionClass, Lambda, type(lambda:1))):
      e = S(e)
      imaybe = e.free_symbols - self.free_symbols
      if not imaybe:
          raise ValueError('expecting a symbol')
      if imaybe and i and i not in imaybe:
          raise ValueError('indicated i not in given expression')
      if len(imaybe) != 1 and not i:
          d = [i for i in imaybe if isinstance(i, Dummy)]
          if len(d) != 1:
              raise ValueError(
                  'not sure what symbol is being used to represent a side')
          i = set(d)
      else:
          i = imaybe
      i = i.pop()
      f = lambda side: e.subs(i, side)
    else:
      f = e
    return self.func(*[f(side) for side in self.args], evaluate=doit)

from sympy.core.relational import Equality
Equality.do = do