To fix the issue you can do I think there's an implicit assumption that flowers of all three colors are represented in the garden. With that in mind, here's how I'd go about coding it using both Python and Haskell interfaces to Z3; because it's just easier to code in those languages than directly in SMTLib.

code :

```
from z3 import *
# Color enumeration
Color, (R, Y, B) = EnumSort('Color', ('R', 'Y', 'B'))
# An uninterpreted function for color assignment
col = Function('col', IntSort(), Color)
# Number of flowers
N = Int('N')
# Helper function to specify distinct valid choices:
def validPick (i1, i2, i3):
return And( Distinct(i1, i2, i3)
, 1 <= i1, 1 <= i2, 1 <= i3
, i1 <= N, i2 <= N, i3 <= N
)
# Helper function to count a given color
def count(pickedColor, flowers):
return Sum([If(col(f) == pickedColor, 1, 0) for f in flowers])
# Get a solver and variables for our assertions:
s = Solver()
f1, f2, f3 = Ints('f1 f2 f3')
# There's at least one flower of each color
s.add(Exists([f1, f2, f3], And(validPick(f1, f2, f3), col(f1) == R, col(f2) == Y, col(f3) == B)))
# You pick three, and at least one of them is red
s.add(ForAll([f1, f2, f3], Implies(validPick(f1, f2, f3), count(R, [f1, f2, f3]) >= 1)))
# You pick three, and at least one of them is yellow
s.add(ForAll([f1, f2, f3], Implies(validPick(f1, f2, f3), count(Y, [f1, f2, f3]) >= 1)))
# For every three flowers you pick, one of them has to be blue
s.add(ForAll([f1, f2, f3], Implies(validPick(f1, f2, f3), count(B, [f1, f2, f3]) == 1)))
# Find a satisfying value of N
print s.check()
print s.model()[N]
# See if there's any other value by outlawing the choice
nVal = s.model()[N]
s.add(N != nVal)
print s.check()
```

```
sat
3
unsat
```

```
print s.sexpr()
```