Fix use of Piecewise in If

Author: bdlucas1

mathics/builtin/arithmetic.py

@@ -372,6 +372,11 @@ def to_sympy(self, expr, **kwargs):
                 sympy_cond = False
         if sympy_cond is None:
             sympy_cond = cond.to_sympy(**kwargs)
+            # See similar code and comment in mathics.builtin.procedural.If
+            # TODO: consider adding .is_Symbol (as with If) so that this
+            # can be used for compilation. I tried that but it invalidated
+            # doctest 1876 in Simplify which depends on ConditionalExpression
+            # not getting rewritten, so may need update that secion of doc?
             if not (sympy_cond.is_Relational or sympy_cond.is_Boolean):
                 return
 

mathics/builtin/procedural.py

@@ -455,7 +455,20 @@ def to_sympy(self, expr, **kwargs):
             sympy_cond = expr.elements[0].to_sympy(**kwargs)
             sympy_true = expr.elements[1].to_sympy(**kwargs)
             sympy_false = expr.elements[2].to_sympy(**kwargs)
-            return sympy.Piecewise((sympy_true, sympy_cond), (sympy_false, True))
+            if sympy_cond is None:
+                return
+            # sympy.Piecewise is is picky as to what type of conds it will accept,
+            # allowing Boolean, Relational, and Symbol (as an honorary Boolean,
+            # accompanied by a comment in the code that this isn't really correct).
+            # Seems an attempt at early typechecking, but maybe too restrictive -
+            # what about Exprs in general that might or might not evaluate to Boolean?
+            # See similar code in mathics.builtin.arithmetic.ConditionalExpression.
+            if (
+                sympy_cond.is_Boolean
+                or sympy_cond.is_Relational
+                or sympy_cond.is_Symbol
+            ):
+                return sympy.Piecewise((sympy_true, sympy_cond), (sympy_false, True))
 
 
 class Interrupt(Builtin):