code: def _solve_inequality(ie, s, linear=False):
      """Return the inequality with s isolated on the left, if possible.
      If the relationship is non-linear, a solution involving And or Or
      may be returned. False or True are returned if the relationship
      is never True or always True, respectively.
  
      If `linear` is True (default is False) an `s`-dependent expression
      will be isolated on the left, if possible
      but it will not be solved for `s` unless the expression is linear
      in `s`. Furthermore, only "safe" operations which do not change the
      sense of the relationship are applied: no division by an unsigned
      value is attempted unless the relationship involves Eq or Ne and
      no division by a value not known to be nonzero is ever attempted.
  
      Examples
      ========
  
      >>> from sympy import Eq, Symbol
      >>> from sympy.solvers.inequalities import _solve_inequality as f
      >>> from sympy.abc import x, y
  
      For linear expressions, the symbol can be isolated:
  
      >>> f(x - 2 < 0, x)
      x < 2
      >>> f(-x - 6 < x, x)
      x > -3
  
      Sometimes nonlinear relationships will be False
  
      >>> f(x**2 + 4 < 0, x)
      False
  
      Or they may involve more than one region of values:
  
      >>> f(x**2 - 4 < 0, x)
      (-2 < x) & (x < 2)
  
      To restrict the solution to a relational, set linear=True
      and only the x-dependent portion will be isolated on the left:
  
      >>> f(x**2 - 4 < 0, x, linear=True)
      x**2 < 4
  
      Division of only nonzero quantities is allowed, so x cannot
      be isolated by dividing by y:
  
      >>> y.is_nonzero is None  # it is unknown whether it is 0 or not
      True
      >>> f(x*y < 1, x)
      x*y < 1
  
      And while an equality (or inequality) still holds after dividing by a
      non-zero quantity
  
      >>> nz = Symbol('nz', nonzero=True)
      >>> f(Eq(x*nz, 1), x)
      Eq(x, 1/nz)
  
      the sign must be known for other inequalities involving > or <:
  
      >>> f(x*nz <= 1, x)
      nz*x <= 1
      >>> p = Symbol('p', positive=True)
      >>> f(x*p <= 1, x)
      x <= 1/p
  
      When there are denominators in the original expression that
      are removed by expansion, conditions for them will be returned
      as part of the result:
  
      >>> f(x < x*(2/x - 1), x)
      (x < 1) & Ne(x, 0)
      """
      from sympy.solvers.solvers import denoms
  
      if s not in ie.free_symbols:
          return ie
      if ie.rhs == s:
          ie = ie.reversed
      if ie.lhs == s and s not in ie.rhs.free_symbols:
          return ie
  
      def classify(ie, s, i):
          # return True or False if ie evaluates when substituting s with
          # i else None (if unevaluated) or NaN (when there is an error
          # in evaluating)
          try:
              v = ie.subs(s, i)
              if v is S.NaN:
                  return v
              elif v not in (True, False):
                  return
              return v
          except TypeError:
              return S.NaN
  
      rv = None
      oo = S.Infinity
      expr = ie.lhs - ie.rhs
      try:
          p = Poly(expr, s)
          if p.degree() == 0:
              rv = ie.func(p.as_expr(), 0)
          elif not linear and p.degree() > 1:
              # handle in except clause
              raise NotImplementedError
      except (PolynomialError, NotImplementedError):
          if not linear:
              try:
                  rv = reduce_rational_inequalities([[ie]], s)
              except PolynomialError:
                  rv = solve_univariate_inequality(ie, s)
              # remove restrictions wrt +/-oo that may have been
              # applied when using sets to simplify the relationship
              okoo = classify(ie, s, oo)
              if okoo is S.true and classify(rv, s, oo) is S.false:
                  rv = rv.subs(s < oo, True)
              oknoo = classify(ie, s, -oo)
              if oknoo is S.true and classify(rv, s, -oo) is S.false:
                  rv = rv.subs(-oo < s, True)
                  rv = rv.subs(s > -oo, True)
              if rv is S.true:
                  rv = (s <= oo) if okoo is S.true else (s < oo)
                  if oknoo is not S.true:
                      rv = And(-oo < s, rv)
          else:
              p = Poly(expr)
  
      conds = []
      if rv is None:
          e = p.as_expr()  # this is in expanded form
          # Do a safe inversion of e, moving non-s terms
          # to the rhs and dividing by a nonzero factor if
          # the relational is Eq/Ne; for other relationals
          # the sign must also be positive or negative
          rhs = 0
          b, ax = e.as_independent(s, as_Add=True)
          e -= b
          rhs -= b
          ef = factor_terms(e)
          a, e = ef.as_independent(s, as_Add=False)
          if (
              a.is_zero != False
              or a.is_negative  # don't divide by potential 0
              == a.is_positive
              is None
              and ie.rel_op  # if sign is not known then
              not in ("!=", "==")
          ):  # reject if not Eq/Ne
              e = ef
              a = S.One
          rhs /= a
          if a.is_positive:
              rv = ie.func(e, rhs)
          else:
              rv = ie.reversed.func(e, rhs)
  
          # return conditions under which the value is
          # valid, too.
          beginning_denoms = denoms(ie.lhs) | denoms(ie.rhs)
          current_denoms = denoms(rv)
          for d in beginning_denoms - current_denoms:
              c = _solve_inequality(Eq(d, 0), s, linear=linear)
              if isinstance(c, Eq) and c.lhs == s:
                  if classify(rv, s, c.rhs) is S.true:
                      # rv is permitting this value but it shouldn't
                      conds.append(~c)
          for i in (-oo, oo):
              if (
                  classify(rv, s, i) is S.true
                  and classify(ie, s, i) is not S.true
              ):
                  conds.append(s < i if i is oo else i < s)
  
      conds.append(rv)
      return And(*conds)
  
imports: import itertools
  from sympy.calculus.util import continuous_domain, periodicity, function_range
  from sympy.core import sympify
  from sympy.core.exprtools import factor_terms
  from sympy.core.relational import Relational, Lt, Ge, Eq
  from sympy.core.symbol import Symbol, Dummy
  from sympy.sets.sets import Interval, FiniteSet, Union, Intersection
  from sympy.core.singleton import S
  from sympy.core.function import expand_mul
  from sympy.functions.elementary.complexes import Abs
  from sympy.logic import And
  from sympy.polys import Poly, PolynomialError, parallel_poly_from_expr
  from sympy.polys.polyutils import _nsort
  from sympy.solvers.solveset import solvify, solveset
  from sympy.utilities.iterables import sift, iterable
  from sympy.utilities.misc import filldedent
label_desc: Use image node_id 9 for calling a global function with example usage: _solve_inequality(ie, s, linear) and returns: And, ie, ie, v, v, S