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> I am using binary decision diagrams. Let's say I am able to determine that
  the intersection of two variables in the BDD, `x` and `y`, are equal to a third
  variable, `z`. How can I replace all occurrences of `x && y` in the BDD with
  `z`?

● To replace all occurrences of x && y with z in a BDD, you need to perform a
  substitution operation. Here are the main approaches:

  1. Variable Substitution

  Most BDD libraries provide a substitution function. You would:
  # Create substitution mapping
  substitution = {(x, y): z}  # Replace x AND y with z
  new_bdd = original_bdd.substitute(substitution)

  2. Compose Operation

  Use the compose operation to replace the conjunction:
  # Create a new BDD representing x && y ↔ z
  equivalence = (x & y).iff(z)
  new_bdd = original_bdd.compose(equivalence)

  3. Manual Traversal and Reconstruction

  For custom implementations, traverse the BDD and rebuild:
  - When you encounter a node where both x and y branches lead to the same result
  - Replace that subgraph with a direct reference to z

  4. Restrict and Replace

  # Restrict the BDD to cases where x && y is true, replace with z
  restricted = original_bdd.restrict({x: True, y: True})
  new_bdd = original_bdd.substitute({x & y: z})

  The exact syntax depends on your BDD library (CUDD, PyBDD, dd, etc.). Which BDD
  library are you using?

> I am writing my own BDD implementation.

● For your own BDD implementation, you'll need to traverse the BDD structure and
  rebuild it. Here's the general approach:

  Core Algorithm

  1. Identify Target Patterns: During traversal, detect when you have a node
     structure representing x && y
  2. Replace with Z: Substitute those subgraphs with references to variable z
  3. Maintain Canonicity: Ensure the resulting BDD remains reduced and canonical

  Implementation Strategy

  def replace_conjunction(bdd_node, x_var, y_var, z_var, memo={}):
      if bdd_node in memo:
          return memo[bdd_node]

      if bdd_node.is_terminal():
          return bdd_node

      # Check if current node represents x && y
      if is_conjunction_pattern(bdd_node, x_var, y_var):
          # Replace with z variable
          result = create_variable_node(z_var)
      else:
          # Recursively process children
          low = replace_conjunction(bdd_node.low, x_var, y_var, z_var, memo)
          high = replace_conjunction(bdd_node.high, x_var, y_var, z_var, memo)
          result = create_node(bdd_node.var, low, high)

      memo[bdd_node] = result
      return result

  def is_conjunction_pattern(node, x_var, y_var):
      # Pattern: node for x with high child being y, low child being 0
      # and y's high child being 1, low child being 0
      if node.var == x_var:
          if node.low.is_zero() and node.high.var == y_var:
              return node.high.low.is_zero() and node.high.high.is_one()
      elif node.var == y_var:
          if node.low.is_zero() and node.high.var == x_var:
              return node.high.low.is_zero() and node.high.high.is_one()
      return False

  Key Considerations

  - Variable Ordering: Maintain your BDD's variable ordering constraints
  - Reduction: Apply reduction rules after substitution to maintain canonical form
  - Shared Nodes: Use memoization to handle shared subgraphs efficiently
  - Edge Cases: Handle cases where x && y appears as part of larger expressions

  What's your current BDD node structure? That would help tailor the pattern
  detection logic.

> Your is_conjunction_pattern will only work when x and y are adjacent (in either
  order) in the BDD graph. What if they are separated by other variables?

● You're absolutely right! My pattern detection was too simplistic. When x and y
  are separated by other variables in the variable ordering, the conjunction
  pattern becomes more complex.

  Proper Conjunction Detection

  You need to check if the paths through the BDD that make both x = 1 and y = 1
  lead to the same result as when z = 1, regardless of intermediate variables.

  def replace_conjunction(bdd_node, x_var, y_var, z_var, memo={}):
      if bdd_node in memo:
          return memo[bdd_node]

      if bdd_node.is_terminal():
          return bdd_node

      # Check if this subgraph represents x && y
      if represents_conjunction(bdd_node, x_var, y_var):
          # Replace entire subgraph with z
          result = create_variable_node(z_var)
      else:
          # Recursively process children
          low = replace_conjunction(bdd_node.low, x_var, y_var, z_var, memo)
          high = replace_conjunction(bdd_node.high, x_var, y_var, z_var, memo)
          result = create_node(bdd_node.var, low, high)

      memo[bdd_node] = result
      return result

  def represents_conjunction(node, x_var, y_var):
      """Check if this BDD subgraph is equivalent to x && y"""
      # Evaluate the subgraph for all combinations of x and y
      assignments = [
          {x_var: False, y_var: False},  # Should be False
          {x_var: False, y_var: True},   # Should be False  
          {x_var: True, y_var: False},   # Should be False
          {x_var: True, y_var: True}     # Should be True
      ]

      expected = [False, False, False, True]

      for i, assignment in enumerate(assignments):
          if evaluate_bdd(node, assignment) != expected[i]:
              return False

      return True

  def evaluate_bdd(node, assignment):
      """Evaluate BDD given variable assignments"""
      if node.is_terminal():
          return node.value

      if node.var in assignment:
          if assignment[node.var]:
              return evaluate_bdd(node.high, assignment)
          else:
              return evaluate_bdd(node.low, assignment)
      else:
          # Variable not assigned - this shouldn't happen in our use case
          # but handle gracefully by trying both branches
          return (evaluate_bdd(node.low, assignment) or
                  evaluate_bdd(node.high, assignment))

  Alternative: Restrict-Based Approach

  A more efficient approach uses BDD restriction operations:

  def represents_conjunction_efficient(node, x_var, y_var):
      """More efficient check using BDD operations"""
      # node should be equivalent to x && y
      # Check: restrict(node, x=0) == 0 and restrict(node, y=0) == 0
      # and restrict(node, x=1, y=1) == 1

      x_false = restrict(node, {x_var: False})
      y_false = restrict(node, {y_var: False})
      both_true = restrict(node, {x_var: True, y_var: True})

      return (is_zero(x_false) and is_zero(y_false) and is_one(both_true))

  This approach correctly handles conjunctions regardless of variable ordering in
  the BDD.

> That still looks like it finds BDD subgraphs that represent `x && y` exactly, not
  subgraphs that _contain_ `x && y` as part of a possibly larger subexpression.

● You're absolutely correct! I was looking for subgraphs that ARE x && y, not
  subgraphs that CONTAIN x && y as a subexpression.

  For replacing x && y with z within larger expressions, you need a different
  approach:

  Shannon Expansion Approach

  Use the Shannon expansion to systematically replace the conjunction:

  def replace_conjunction_in_expression(bdd_node, x_var, y_var, z_var, memo={}):
      if bdd_node in memo:
          return memo[bdd_node]

      if bdd_node.is_terminal():
          return bdd_node

      # Shannon expansion: f = x̄·f|x=0 + x·f|x=1
      f_x0 = restrict(bdd_node, {x_var: False})
      f_x1 = restrict(bdd_node, {x_var: True})

      # Further expand f|x=1 with respect to y
      f_x1_y0 = restrict(f_x1, {y_var: False})
      f_x1_y1 = restrict(f_x1, {y_var: True})

      # Now we have: f = x̄·f|x=0 + x·ȳ·f|x=1,y=0 + x·y·f|x=1,y=1
      # Replace x·y term with z

      # Recursively process the parts
      part_x0 = replace_conjunction_in_expression(f_x0, x_var, y_var, z_var, memo)
      part_x1_y0 = replace_conjunction_in_expression(f_x1_y0, x_var, y_var, z_var, memo)
      part_x1_y1 = replace_conjunction_in_expression(f_x1_y1, x_var, y_var, z_var, memo)

      # Reconstruct: x̄·part_x0 + x·ȳ·part_x1_y0 + z·part_x1_y1
      result = ite(x_var,
                  ite(y_var, part_x1_y1, part_x1_y0),
                  part_x0)

      # But we want to replace x·y with z, so:
      result = ite(z_var, part_x1_y1,
                  ite(x_var,
                     ite(y_var, zero_node(), part_x1_y0),
                     part_x0))

      memo[bdd_node] = result
      return result

  Alternative: Composition Approach

  Create the equivalence relation and compose:

  def replace_conjunction_by_composition(original_bdd, x_var, y_var, z_var):
      # Create BDD representing: (x ∧ y) ↔ z
      # This is: (x ∧ y ∧ z) ∨ (¬(x ∧ y) ∧ ¬z)
      # Which is: (x ∧ y ∧ z) ∨ ((¬x ∨ ¬y) ∧ ¬z)

      xy_and_z = and_bdd(and_bdd(x_var, y_var), z_var)
      not_xy = or_bdd(not_bdd(x_var), not_bdd(y_var))
      not_xy_and_not_z = and_bdd(not_xy, not_bdd(z_var))

      equivalence = or_bdd(xy_and_z, not_xy_and_not_z)

      # Now compose this with the original BDD
      # This constrains the result to satisfy the equivalence
      return and_bdd(original_bdd, equivalence)

  The Shannon expansion approach directly transforms the BDD structure, while the
  composition approach adds constraints. Which approach fits better with your BDD
  operations?

> Are the recursive calls in the Shannon expansion approach necessary? The
  restrict calls should remove all references to x and y from the BDD, if I.
  understand it right.

● You're absolutely right! The recursive calls are unnecessary and actually
  counterproductive.

  When you do restrict(f_x1_y1, {y_var: True}), you're restricting f|x=1 with
  y=1, which gives you f|x=1,y=1. This result has no references to x or y
  variables - they've been eliminated by the restriction operations.

  The corrected approach should be:

  def replace_conjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Shannon expansion to isolate x && y term
      f_x0 = restrict(bdd_node, {x_var: False})           # No x or y refs after restriction
      f_x1_y0 = restrict(bdd_node, {x_var: True, y_var: False})  # No x or y refs
      f_x1_y1 = restrict(bdd_node, {x_var: True, y_var: True})   # No x or y refs

      # Reconstruct: x̄·f|x=0 + x·ȳ·f|x=1,y=0 + z·f|x=1,y=1
      # (replacing the x·y term with z)
      result = ite(z_var, f_x1_y1,
                  ite(x_var,
                     ite(y_var, zero_node(), f_x1_y0),
                     f_x0))

      return result

  No recursive calls needed since the restricted BDDs (f_x0, f_x1_y0, f_x1_y1)
  already have x and y eliminated. Thanks for catching that inefficiency!

  The key insight is that restriction operations completely eliminate the
  restricted variables from the resulting BDD subgraph.

> There might still be references to y in f_x0, but I think that will be okay,
  based on how everything is combined back to together in the ite calls at the
  end.

● You're absolutely correct! There can still be references to y in f_x0 (the
  case where x=False), and that's perfectly fine.

  When we reconstruct with:
  result = ite(z_var, f_x1_y1,
              ite(x_var,
                 ite(y_var, zero_node(), f_x1_y0),
                 f_x0))

  The logic is:
  - If z=True: use f_x1_y1 (the behavior when x=True, y=True)
  - If z=False and x=True and y=True: use zero_node() (since we replaced x∧y
    with z)
  - If z=False and x=True and y=False: use f_x1_y0
  - If z=False and x=False: use f_x0 (which may still depend on y)

  The y references in f_x0 are handled correctly because when x=False, the
  value of y still matters for the overall expression - we only replaced the
  specific x∧y conjunction with z, not all references to y.

  So the algorithm correctly preserves the semantics: x∧y gets replaced with z,
  but other occurrences of y (like when x=False) remain unchanged.

> Yes that sounds right to me. I will give this a shot!

● Great! That approach should work well. The Shannon expansion method is elegant
  because it cleanly separates out the x && y term that you want to replace while
  preserving all other dependencies on x and y in the expression.

  Good luck with the implementation!

> Can you create a similar Shannon expansion method for replacing the union
  `x || y` with `z`?

● Yes! Here's the Shannon expansion approach for replacing x || y with z:

  def replace_disjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Shannon expansion to isolate x || y term
      f_x0_y0 = restrict(bdd_node, {x_var: False, y_var: False})  # No x or y refs
      f_x0_y1 = restrict(bdd_node, {x_var: False, y_var: True})   # No x or y refs  
      f_x1 = restrict(bdd_node, {x_var: True})                    # No x refs, may have y refs

      # Original: f = x̄·ȳ·f|x=0,y=0 + x̄·y·f|x=0,y=1 + x·f|x=1
      # We want to replace x||y with z:
      # x||y is true when: (x=0,y=1) OR (x=1,y=0) OR (x=1,y=1)
      # x||y is false when: (x=0,y=0)

      # Reconstruct: z·f|x||y=1 + z̄·f|x=0,y=0
      # But f|x||y=1 needs to handle the cases where x||y was true

      # For the disjunction case, we need the value when x||y is true
      # This is trickier - we need f|x=0,y=1 unified with f|x=1
      # Since f|x=1 may still depend on y, we need to be more careful

      f_x1_y0 = restrict(bdd_node, {x_var: True, y_var: False})   # No x or y refs
      f_x1_y1 = restrict(bdd_node, {x_var: True, y_var: True})    # No x or y refs

      # The behavior when x||y is true should combine:
      # - f|x=0,y=1 (case where x=false, y=true)  
      # - f|x=1,y=0 (case where x=true, y=false)
      # - f|x=1,y=1 (case where x=true, y=true)
      # But these might have different values, so we need to reconstruct carefully

      # Actually, let's think differently:
      # f = z̄·(x̄·ȳ·f|x=0,y=0) + z·(x̄·y·f|x=0,y=1 + x·ȳ·f|x=1,y=0 + x·y·f|x=1,y=1)
      # But when z=true, we don't want the x,y structure - we want a single value

      # The correct approach: when z replaces x||y, all the "true" cases should behave the same
      # We need to pick a representative value - let's use the x=1,y=1 case
      f_xy_true_representative = f_x1_y1

      result = ite(z_var, f_xy_true_representative, f_x0_y0)

      return result

  Wait, this approach has a problem - when x||y is true in different ways, the
  function might have different values. Let me revise:

  def replace_disjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Shannon expansion
      f_x0_y0 = restrict(bdd_node, {x_var: False, y_var: False})  # x||y = false
      f_x0_y1 = restrict(bdd_node, {x_var: False, y_var: True})   # x||y = true
      f_x1_y0 = restrict(bdd_node, {x_var: True, y_var: False})   # x||y = true  
      f_x1_y1 = restrict(bdd_node, {x_var: True, y_var: True})    # x||y = true

      # Reconstruct: z̄·f|x=0,y=0 + z·(x̄·y·f|x=0,y=1 + x·ȳ·f|x=1,y=0 + x·y·f|x=1,y=1)
      # The z=true case combines the three cases where x||y was true
      xy_true_cases = ite(x_var,
                         ite(y_var, f_x1_y1, f_x1_y0),
                         f_x0_y1)  # Note: when x=false in this branch, y must be true

      result = ite(z_var, xy_true_cases, f_x0_y0)

      return result

  This preserves the original behavior in all the cases where x||y was true, just
  routes them through z instead.

> xy_true_cases will reintroduce x and y as variables in the BDD, which we are
  trying to remove in that case. Would it work to just take the union of f_x1_y1,
  f_x1_y0, and f_x0_y1?

● You're absolutely right! The ite calls reintroduce x and y variables, defeating
  the purpose of the replacement.

  Taking the union (OR) of the three cases is the correct approach:

  def replace_disjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Shannon expansion
      f_x0_y0 = restrict(bdd_node, {x_var: False, y_var: False})  # x||y = false
      f_x0_y1 = restrict(bdd_node, {x_var: False, y_var: True})   # x||y = true
      f_x1_y0 = restrict(bdd_node, {x_var: True, y_var: False})   # x||y = true  
      f_x1_y1 = restrict(bdd_node, {x_var: True, y_var: True})    # x||y = true

      # Union of all cases where x||y was true
      xy_true_union = or_bdd(or_bdd(f_x0_y1, f_x1_y0), f_x1_y1)

      # Reconstruct with z
      result = ite(z_var, xy_true_union, f_x0_y0)

      return result

  This way, when z=True, we get the union of all behaviors that occurred when x||y
  was true, and when z=False, we get the behavior when x||y was false. The
  resulting BDD has no references to x or y in the z=True branch.

  Perfect insight!

> I'm getting a strange result with this last function. When bdd_node represents
  `x`, x_var is `x`, y_var is `y`, and z_var is `y`, then final result is `y`.
  That is, if I want to use the function to replace occurrences of `x || y` with
  `y`, in the BDD representing the function `x`, the result I get is `y`. This
  doesn't seem right. `x || y` doesn't occur in the function `x`, so it seems like
  no substitution should occur.

● You're absolutely right! That's a bug in the logic. When the BDD represents just
  x, there's no occurrence of x || y to replace, so the result should still be x.

  Let's trace through what's happening:

  For BDD representing x:
  - f_x0_y0 = restrict(x, {x: False, y: False}) = False
  - f_x0_y1 = restrict(x, {x: False, y: True}) = False
  - f_x1_y0 = restrict(x, {x: True, y: False}) = True
  - f_x1_y1 = restrict(x, {x: True, y: True}) = True

  So:
  - xy_true_union = False || True || True = True
  - result = ite(y, True, False) = y

  The problem is that we're unconditionally replacing any dependence on the
  combination of x and y values with z, even when the original function doesn't
  actually contain x || y as a logical subexpression.

  We need to first check if the function actually contains x || y as a
  subexpression:

  def replace_disjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Check if x||y actually appears as a subexpression
      if not contains_disjunction(bdd_node, x_var, y_var):
          return bdd_node  # No replacement needed

      # Rest of the Shannon expansion...

  Where contains_disjunction would check if there's actually an x || y pattern to
  replace. But this is getting complex - maybe we need a different approach
  entirely?

> I added a check at the end to see if the result was smaller than the input, on
  the assumption that valid substitutions should only make the result smaller (in
  terms of the number of internal nodes in the BDD). This worked for other cases,
  but in this case, the result has the same size as the input. Technically that's
  "not smaller", but this size check seems brittle.

● You're right that the size check is brittle. The fundamental issue is that
  Shannon expansion always decomposes the function into all possible combinations
  of x and y values, regardless of whether x || y actually appears as a meaningful
  subexpression.

  A better approach might be to check if the replacement actually preserves the
  function's semantics. Here's a more robust validation:

  def replace_disjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Perform the Shannon expansion replacement
      f_x0_y0 = restrict(bdd_node, {x_var: False, y_var: False})
      f_x0_y1 = restrict(bdd_node, {x_var: False, y_var: True})
      f_x1_y0 = restrict(bdd_node, {x_var: True, y_var: False})
      f_x1_y1 = restrict(bdd_node, {x_var: True, y_var: True})

      xy_true_union = or_bdd(or_bdd(f_x0_y1, f_x1_y0), f_x1_y1)
      candidate_result = ite(z_var, xy_true_union, f_x0_y0)

      # Validate: check if this replacement preserves semantics
      if is_valid_disjunction_replacement(bdd_node, candidate_result, x_var, y_var, z_var):
          return candidate_result
      else:
          return bdd_node  # No valid replacement found

  def is_valid_disjunction_replacement(original, replacement, x_var, y_var, z_var):
      """Check if replacement correctly substitutes x||y with z"""
      # The replacement should satisfy: original[x||y := z] ≡ replacement
      # Test on several random assignments or use BDD equivalence checking

      # Quick test: when z matches x||y, should they be equivalent?
      test_cases = [
          ({x_var: False, y_var: False, z_var: False}, True),  # z = x||y
          ({x_var: False, y_var: True, z_var: True}, True),   # z = x||y
          ({x_var: True, y_var: False, z_var: True}, True),   # z = x||y
          ({x_var: True, y_var: True, z_var: True}, True),    # z = x||y
      ]

      for assignment, should_match in test_cases:
          if should_match:
              orig_val = evaluate_bdd(original, assignment)
              repl_val = evaluate_bdd(replacement, assignment)
              if orig_val != repl_val:
                  return False

      return True

  This semantic validation approach is more reliable than size heuristics.

> What would "BDD equivalence" checking look like?

● BDD equivalence checking is actually very straightforward - it's one of the
  main advantages of using BDDs!

  Structural Equivalence

  If your BDDs are canonical (reduced and follow a consistent variable ordering),
  then:

  def are_equivalent(bdd1, bdd2):
      return bdd1 is bdd2  # Pointer equality!

  This works because canonical BDDs have the property that equivalent Boolean
  functions always have identical BDD representations.

  If Not Canonical

  If your BDDs might not be canonical, you can check structural equality:

  def are_equivalent(bdd1, bdd2):
      if bdd1 is bdd2:
          return True

      if bdd1.is_terminal() and bdd2.is_terminal():
          return bdd1.value == bdd2.value

      if bdd1.is_terminal() or bdd2.is_terminal():
          return False

      return (bdd1.var == bdd2.var and
              are_equivalent(bdd1.low, bdd2.low) and
              are_equivalent(bdd1.high, bdd2.high))

  Using XOR for Equivalence

  Another elegant approach uses the fact that f ≡ g iff f ⊕ g ≡ 0:

  def are_equivalent(bdd1, bdd2):
      xor_result = xor_bdd(bdd1, bdd2)
      return xor_result.is_zero()

  For Your Validation

  In your validation function:

  def is_valid_disjunction_replacement(original, replacement, x_var, y_var, z_var):
      # Create constraint: z ≡ (x || y)
      constraint = iff_bdd(z_var, or_bdd(x_var, y_var))

      # Under this constraint, original and replacement should be equivalent
      constrained_original = and_bdd(original, constraint)
      constrained_replacement = and_bdd(replacement, constraint)

      return are_equivalent(constrained_original, constrained_replacement)

  The canonical property makes BDD equivalence checking extremely efficient -
  constant time for canonical BDDs!

> Is there a similar validation function we can use for the earlier shannon
  expansion for intersection?

● Yes! Here's the validation function for the conjunction replacement:

  def is_valid_conjunction_replacement(original, replacement, x_var, y_var, z_var):
      """Check if replacement correctly substitutes x&&y with z"""
      # Create constraint: z ≡ (x && y)
      constraint = iff_bdd(z_var, and_bdd(x_var, y_var))

      # Under this constraint, original and replacement should be equivalent
      constrained_original = and_bdd(original, constraint)
      constrained_replacement = and_bdd(replacement, constraint)

      return are_equivalent(constrained_original, constrained_replacement)

  And you'd use it in your conjunction replacement function:

  def replace_conjunction_in_expression(bdd_node, x_var, y_var, z_var):
      if bdd_node.is_terminal():
          return bdd_node

      # Shannon expansion
      f_x0 = restrict(bdd_node, {x_var: False})
      f_x1_y0 = restrict(bdd_node, {x_var: True, y_var: False})
      f_x1_y1 = restrict(bdd_node, {x_var: True, y_var: True})

      # Reconstruct with z replacing x&&y
      candidate_result = ite(z_var, f_x1_y1,
                            ite(x_var,
                               ite(y_var, zero_node(), f_x1_y0),
                               f_x0))

      # Validate the replacement
      if is_valid_conjunction_replacement(bdd_node, candidate_result, x_var, y_var, z_var):
          return candidate_result
      else:
          return bdd_node  # No valid replacement

  The validation logic is identical - we constrain both the original and
  replacement functions with z ≡ (x && y) and check if they're equivalent under
  that constraint. This should correctly reject cases like replacing x && y with
  z in the function x, since x under the constraint z ≡ (x && y) is not
  equivalent to z under the same constraint.

> I have updated these functions to also accept negated variables for x_var and
  y_var.

● That's a great extension! Supporting negated variables makes the replacement
  functions much more flexible. You can now handle patterns like:

  - ¬x && y with z
  - x && ¬y with z
  - ¬x && ¬y with z
  - ¬x || y with z
  - x || ¬y with z
  - ¬x || ¬y with z

  This significantly expands the kinds of logical equivalences you can exploit
  in BDD optimization. Are you representing negated variables as separate
  objects, or do you have a way to encode the negation flag within the variable
  representation?

  Also, did you need to adjust the validation functions to handle the negated
  cases, or do they work automatically with your negated variable
  representation?

> I am representing negated variables as separate objects. This project is in
  Rust, and there is an enum with Positive and Negative variants, each of which
  wraps a variable.

● That's a clean design! Using an enum like:

  enum Literal {
      Positive(Variable),
      Negative(Variable),
  }

  makes the negation explicit and type-safe. This approach probably made
  extending the Shannon expansion functions straightforward - you can just
  evaluate the literal (positive or negative) when doing the restrictions, and
  the validation functions should work automatically since they operate on the
  literal values rather than assuming positive variables.

  Does your restrict function handle Literal types directly, or do you convert
  them when building the restriction map?

> The restrict function takes in an iterator of Literals. (There is a helper
  function called restrict_one that restricts a single variable assignment.)

  re your other question, I _think_ the validation functions work as-is,
  because `ite` takes in a BDD node as a conditional, not a variable or
  literal, so I can pass in the simple BDD node representing a negated
  variable for the negated case. Same for the `iff`, `and`, and `or` functions
  that I use to build up the validation predicate.

  Unfortunately, there are still test failures! I cannot tell if it's because
  there is an issue with these replace_* functions, or because I am using
  them wrong. I can give more detail about what the variables represent,
  and what I am using the replace functions for, if that would help diagnose.

● Yes, providing more context about what the variables represent and how
  you're using the replace functions would definitely help diagnose the issues!

  A few questions that might help narrow down the problem:

  1. What do the variables represent in your BDD application?
  2. How are you calling the replace functions - are you chaining multiple
     replacements, or doing single replacements?
  3. What kind of test failures are you seeing - incorrect results, panics,
     or something else?
  4. Do you have any example cases where the replacement is clearly wrong
     that we could trace through?

  Also, you're right that the validation functions should work with negated
  variables since they just create BDD nodes representing the literals. But
  there might be edge cases we're missing.

  The Shannon expansion approach assumes that we're doing a "global" replacement
  throughout the entire BDD. If you're trying to do more targeted replacements
  (like only replacing in certain contexts), that could also cause issues.

  What's the broader context of what you're trying to achieve with these
  replacements?

> I am working on a Python type checker, and am working on infering
  specializations of type variables in generic functions. I am using BDDs to
  represent the constraints on the typevars in a particular call. Each BDD
  variable represents a range of types that a single typevar can be specialized
  to. So each variable has the form `LowerBound ≤ TypeVariable ≤ UpperBound`. We
  are treating types as sets of runtime values, so the ≤ comparison function is
  a simple subset check. That means that types and type variables form a poset.

  I am using these replacement functions to try to simplify BDDs (constraint
  sets) that have multiple clauses constraining the same typevar. The bounds
  might be related to each other. For instance, if there are types
  `A ≤ B ≤ C ≤ D`, and a typevar `T`, then the constraint `A ≤ T ≤ D` (which
  we might call BDD variable `x`) completely subsumes the constraint `B ≤ T ≤ C`
  (which we can call BDD variable `y`). That means the implication `y -> x`
  should hold, and so we should be able to replace `x && y` in any BDD with `y`.
  (The intersection of those two constraints should be the "smaller"
  constraint.)

● Ah, this provides crucial context! The issue is with constraint subsumption,
  not logical equivalence.

  If x represents A ≤ T ≤ D and y represents B ≤ T ≤ C, where A ≤ B ≤ C ≤ D,
  then:
  - y implies x (the narrower constraint implies the broader one)
  - x && y should simplify to just y (the intersection is the narrower
    constraint)

  But the Shannon expansion functions I provided assume logical equivalence
  (x && y ≡ z), not implication (y → x, therefore x && y ≡ y).

● For constraint subsumption, you need different replacement functions:

  // When y implies x (y is more restrictive), replace x && y with y
  fn replace_subsumed_conjunction(bdd_node, broader_var, narrower_var) {
      // Shannon expansion
      let f_broader0 = restrict(bdd_node, [(broader_var, false)]);
      let f_broader1_narrower0 = restrict(bdd_node, [(broader_var, true), (narrower_var, false)]);
      let f_broader1_narrower1 = restrict(bdd_node, [(broader_var, true), (narrower_var, true)]);

      // When both are true, result should be same as when just narrower is true
      // When broader is false, narrower must also be false (due to implication)
      ite(narrower_var, f_broader1_narrower1, f_broader0)
  }

● The validation function should check for implication, not equivalence:

  fn is_valid_subsumption_replacement(original, replacement, broader_var, narrower_var) {
      // Create constraint: narrower → broader (if narrower then broader)
      let implication = or_bdd(not_bdd(narrower_var), broader_var);

      // Under this constraint, original and replacement should be equivalent
      let constrained_original = and_bdd(original, implication);
      let constrained_replacement = and_bdd(replacement, implication);

      are_equivalent(constrained_original, constrained_replacement)
  }

  This explains your test failures - you're using equivalence-based replacement
  for implication-based constraints!

> There are also constraints that do form the logical equivalence form. For
  instance, with the same types and typevar, given constraints `A ≤ T ≤ C`
  (variable `x1`) and `B ≤ T ≤ D` (variable `x2`), the intersection
  (`x1 && x2`) should simplify to `B ≤ T ≤ C`, which we could represent with
  a new variable `x3`. I would use the previous replacement functions to
  replace all occurrences of `x1 && x2` with `x3`.

● Exactly! You have two different types of simplifications:

  1. Subsumption (Implication-based)

  When one constraint completely contains another:
  - A ≤ T ≤ D (broader) and B ≤ T ≤ C (narrower), where A ≤ B ≤ C ≤ D
  - narrower → broader, so broader && narrower ≡ narrower
  - Use the subsumption replacement functions I mentioned above

  2. Intersection (Equivalence-based)

  When two constraints partially overlap:
  - A ≤ T ≤ C and B ≤ T ≤ D intersect to form B ≤ T ≤ C (assuming
    A ≤ B ≤ C ≤ D)
  - x1 && x2 ≡ x3 where x3 represents the intersection constraint
  - Use the original Shannon expansion functions for logical equivalence

  So you need both types of replacement functions in your toolkit! The key is
  correctly identifying which situation you're in:

  - Subsumption: One constraint's bounds completely contain the other's bounds
  - Intersection: Neither constraint subsumes the other, but they have a
    non-empty intersection

  Are you currently trying to handle both cases, or focusing on one type? The
  test failures might be from mixing up which replacement function to use for
  which type of constraint relationship.

> Can the subsumption case not also be handled as an intersection? As you
  point out, when narrower → broader, then broader && narrower = narrower. Is
  there a problem since the replacement includes one of the variables being
  replaced?

  Further complicating things is that there can be negative constraints, such
  as `¬(A ≤ T ≤ D)`. (In Python, there is a universal bottom type called
  `Never` and a universal top type called `object`.) So I also have to
  consider simplifications involving positive and/or negative constraints.
  For instance, the intersection of `A ≤ T ≤ B` and `¬(C ≤ T ≤ D)` should be
  `A ≤ T ≤ B`. (The ranges of the two constraints do not overlap, so the
  negative constraint does not remove any of the types that are included in
  the positive constraint.) This is why I added support for negative
  variables in the Shannon expansion functions — to allow me to represent
  `x4 && ¬x5 = x4`. (Or alternatively, `(x4 → x5) → (x4 && x5 = x4)`, if your
  hunch is correct that I have to use the subsumption function for that
  case.)

● You're absolutely right! The subsumption case can be handled as
  intersection where one operand equals the result. There's nothing
  fundamentally wrong with x1 && x2 ≡  x2 as a replacement pattern.

  The Shannon expansion approach should handle this correctly:

  // For x1 && x2 ≡ x2 (where x2 is the narrower constraint)
  f_x1_0 = restrict(bdd, {x1: false});           // When broader constraint fails
  f_x1_1_x2_0 = restrict(bdd, {x1: true, x2: false}); // Broader true, narrower false  
  f_x1_1_x2_1 = restrict(bdd, {x1: true, x2: true});  // Both true

  // Reconstruct: x2 ? f_x1_1_x2_1 : (x1 ? f_x1_1_x2_0 : f_x1_0)
  // But since x2 → x1, when x2=true we know x1=true, and when x2=false and x1=true, it's impossible
  result = ite(x2, f_x1_1_x2_1, f_x1_0);

  Your negative constraint examples are perfect use cases for the Shannon
  expansion with negated literals:

  - x4 && ¬x5 ≡ x4 (when ranges don't overlap)
  - This is exactly the pattern the functions should handle

  The issue might be in the validation functions - they need to account for
  the constraint relationships (subsumption, non-overlapping ranges) that make
  these equivalences valid in your domain, even if they wouldn't be valid in
  pure Boolean logic.

  Are you seeing failures in cases where the mathematical constraint
  relationship is correct, but the BDD replacement gives wrong results?

> Yes

● The problem is likely that the validation functions are checking for pure
  Boolean equivalence, but your equivalences are only valid under the domain
  constraints of your type system.

  For example, x1 && x2 ≡ x2 is not true in general Boolean logic, but it IS
  true when you have the constraint that x2 → x1 (narrower implies broader).

  You need domain-aware validation:

  fn is_valid_constraint_replacement(original, replacement, constraints) {
      // Add your domain constraints to both sides
      let constrained_original = and_bdd(original, constraints);
      let constrained_replacement = and_bdd(replacement, constraints);

      are_equivalent(constrained_original, constrained_replacement)
  }

  // Usage for subsumption case
  fn validate_subsumption(original, replacement, broader_var, narrower_var) {
      let implication = or_bdd(not_bdd(narrower_var), broader_var); // narrower → broader
      is_valid_constraint_replacement(original, replacement, implication)
  }

  // Usage for non-overlapping negative constraint
  fn validate_non_overlapping(original, replacement, pos_var, neg_var) {
      let non_overlap = not_bdd(and_bdd(pos_var, neg_var)); // ¬(pos ∧ neg)
      is_valid_constraint_replacement(original, replacement, non_overlap)
  }

  The current validation functions assume the replacement should work in all
  possible Boolean assignments, but your replacements are only valid under
  your type constraint assumptions.

  Try replacing the validation with domain-aware versions that include your
  constraint relationships. That should fix the test failures.

> x1 && x2 = x2 is not true for an arbitrary x1 and x2. But if we have
  established that x1 and x2 represent constraints where one consumes the
  other, then it _does_ hold for those particular variables, no matter which
  BDD they appear in. Does that not mean that we can treat x1 && x2 = x2 as
  an equivalence for those particular variables?

● You're absolutely correct! If you've globally established that x2 → x1
  (e.g., through static analysis of your constraint bounds), then
  x1 && x2 ≡ x2 should hold everywhere those variables appear, in any BDD.

  This means the Shannon expansion approach should work correctly, and you
  shouldn't need domain-aware validation - the equivalence is universal for
  those specific variables.

  If you're still getting test failures with this approach, the issue is likely
  in the implementation details rather than the conceptual approach. Some things
  to check:

  1. Variable ordering: Are x1 and x2 in the expected order in your BDD variable
     ordering?
  2. Edge cases: What happens when the BDD doesn't actually contain both
     variables? (Should return unchanged)
  3. Negated variables: Are you handling the cases with negated literals
     correctly in the Shannon expansion?
  4. Implementation bugs: Are the restrict, ite, and_bdd, etc. functions working
     correctly?

  Could you share a specific failing test case? Something like:
  - Input BDD representing what Boolean function
  - Which variables you're trying to replace (x1 && x2 → x2)
  - What you get vs. what you expect

  That would help pinpoint whether it's a conceptual issue or an implementation
  bug.

> Sure! I have a constraint `(B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated)`. I have
  established that `B ≤ C` and that `Unrelated` is not comparable to `B` or
  `C`. Therefore the two ranges are disjoint, and so this falls into the
  positive/negative case that we have been discussing. I can assume that
  `(B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated) = (B ≤ T ≤ C)`, and also that
  `¬(B ≤ T ≤ C) ∧ (T ≤ Unrelated) = (T ≤ Unrelated)`.

  When I use my Shannon expansion method on the first equivalence, it correctly
  substitutes, and returns `(B ≤ T ≤ C)`. However, I don't know in advance
  which equivalences will be needed when specifying any particular BDD, so I
  then also try to substitute the second equivalence. The lhs does not appear
  in the function `(B ≤ T  ≤ C)`, but the Shannon expansion method still tries
  to substitute it, producing `(B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated)` as a final
  result. In effect, it undid the substitution that we correctly applied in the
  first call.

● Ah! This is the exact problem we identified earlier with unconditional Shannon
  expansion. The second replacement is incorrectly applying because
  ¬(B ≤ T ≤ C) ∧ (T ≤ Unrelated) doesn't actually appear in the function
  (B ≤ T ≤ C).

  The Shannon expansion method always decomposes the function based on all
  combinations of the variables, even when the target pattern isn't present. In
  your case:

  For function (B ≤ T ≤ C) with variables pos = (B ≤ T ≤ C) and
  neg = (T ≤ Unrelated):
  - f_pos1_neg0 = restrict(pos, {pos: true, neg: false}) = true
  - f_pos0_neg1 = restrict(pos, {pos: false, neg: true}) = false
  - Reconstructed: ite(neg, false, ite(pos, true, false)) = pos ∧ ¬neg

  This artificially introduces the neg variable even though it wasn't in the
  original function!

  Solution: You need to check if the pattern actually exists before applying
  the replacement:

  fn replace_conjunction_if_present(bdd_node, x_var, y_var, z_var) {
      // Check if both x_var and y_var actually appear in the BDD
      if !contains_variable(bdd_node, x_var) || !contains_variable(bdd_node, y_var) {
          return bdd_node;  // Pattern can't exist, no replacement needed
      }

      // Proceed with Shannon expansion...
  }

  fn contains_variable(bdd_node, var) -> bool {
      // Traverse BDD to check if variable appears
      // (memoized traversal)
  }

  This prevents the "false positive" replacements that undo previous work.

> Isn't that what we tried to fix by introducing the semantic validation
  function? Do we also need to verify that all variables exist in the function
  being simplified? Do we need both that _and_ semantic validation? Or is there
  maybe a problem with the semantic validation functions we're using?

● You're right to question this! Let's think through what each approach is
  supposed to catch:

  Semantic validation should catch the case where the replacement doesn't
  preserve semantics - but the issue is that in your second replacement attempt,
  the Shannon expansion is creating a semantically different function by
  introducing new variables.

  Let's trace what happens with your example:

  Original: (B ≤ T ≤ C)
  Trying to replace: ¬(B ≤ T ≤ C) ∧ (T ≤ Unrelated) → (T ≤ Unrelated)

  Shannon expansion produces: (B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated)

  The semantic validation should reject this because:
  - Original under constraint: (B ≤ T ≤ C) ∧ [constraint about disjoint ranges]
  - Replacement under constraint: (B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated) ∧ [constraint]

  These are not equivalent - the replacement has introduced an extra variable.

  If your semantic validation is not catching this, there might be a bug in how
  you're constructing the constraint or checking equivalence. Can you share how
  you're implementing the semantic validation for this non-overlapping case?

  The issue might be that the constraint ¬(pos ∧ neg) allows cases where neg is
  false, which makes the replacement appear valid when it shouldn't be.

> As suggested earlier, since this is substituting an intersection, I am using
  `¬(B ≤ T ≤ C) ∧ (T ≤ Unrelated) <=> (T ≤ Unrelated)` as the validation
  constraint. My BDD implementation claims that simplifies to
  `¬(T ≤ Unrelated) ∨ ¬(B ≤ T ≤ C)`. (That seems correct to me.) So with an
  input of `(B ≤ T ≤ C)`, the Shannon expansion (before checking the validity
  constraint) is `(B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated)`. Combining the validity
  constraint with the input yields `(B ≤ T ≤ C) ∧ ¬(T ≤ Unrelated)`, and 
  combining the validity constraint with the Shannon expansion yields the same.

● I see the issue! Your equivalence constraint is backwards. You're using:

  ¬(B ≤ T ≤ C) ∧ (T ≤ Unrelated) ≡ (T ≤ Unrelated)

  But this is saying "when the positive constraint is false AND the negative is
  true, then just the negative is true." However, you're trying to replace
  ¬pos ∧ neg with neg everywhere, not just when this equivalence holds.

  The validation is passing because:
  - Original: pos
  - Shannon result: pos ∧ ¬neg
  - Constraint: ¬neg ∨ ¬pos (your equivalence constraint)

  Under the constraint ¬neg ∨ ¬pos:
  - pos ∧ (¬neg ∨ ¬pos) = pos ∧ ¬neg (since pos ∧ ¬pos = false)
  - (pos ∧ ¬neg) ∧ (¬neg ∨ ¬pos) = pos ∧ ¬neg

  So they're "equivalent" under the constraint, but that's because the
  constraint is forcing neg = false whenever pos = true!

  The real issue: Shannon expansion unconditionally introduces variables that
  shouldn't be there. You need the variable existence check in addition to
  semantic validation:

  if !contains_variable(bdd_node, x_var) || !contains_variable(bdd_node, y_var) {
      return bdd_node;  // Skip replacement - pattern doesn't exist
  }

  The semantic validation alone isn't sufficient because it can't distinguish
  between "replacement is valid" and "constraint forces equivalence by
  eliminating cases."

> Okay I will give that a try!

● Sounds good! The variable existence check should prevent the Shannon
  expansion from introducing variables that weren't originally present in the
  BDD, which is what's causing your replacements to interfere with each other.

  Let me know how it works out - this should fix the issue where the second
  replacement was undoing the first one.