Not a proof, but if you look at it as a satisfiability problem, Z3 (SMT solver) can be of some help here:
from z3 import *
o = Optimize()
# Let's number the pins from 0 to 9, it's much easier that way
# A = 0, B = 1, C = 2, D = 3, E = 4, F = 5, G = 6, H = 7, I = 8,
# J = 9
pins = [Bool("%d" % i) for i in range(0, 10)]
# Since there are 15 equilateral triangles, let's
# add them as constraints to be solved
o.add(And(Or(pins[0], pins[1], pins[2]),
Or(pins[1], pins[3], pins[4]),
Or(pins[2], pins[4], pins[5]),
Or(pins[3], pins[6], pins[7]),
Or(pins[4], pins[7], pins[8]),
Or(pins[5], pins[8], pins[9]),
Or(pins[4], pins[1], pins[2]),
Or(pins[7], pins[3], pins[4]),
Or(pins[8], pins[4], pins[5]),
Or(pins[0], pins[6], pins[9]),
Or(pins[1], pins[6], pins[8]),
Or(pins[2], pins[7], pins[9]),
Or(pins[0], pins[3], pins[5]),
Or(pins[3], pins[2], pins[8]),
Or(pins[5], pins[1], pins[7])))
# minimize takes a variable, so let's pass
# in sum since we can't pass the entire list
o.minimize(Sum(pins))
# check for satisfiability and
# get the pins to be removed if it
# is satisfiable
if o.check() == sat:
# Get the model
m = o.model()
# Print the number of pins to be removed
print("The number of pins to be removed is %s" % str((m.evaluate(Sum(pins)))))
# Print the pins to be removed
print("The pins to be removed are: ", end="")
for i in range(0, 10):
if m.evaluate(pins[i]) == True:
print("%d" % i, end=" ")
print()
else:
print("The problem is unsatisfiable")
This spits out:
The number of pins to be removed is 4
The pins to be removed are: 0 4 7 8
Which is the same as A, E, H, and I if you map 0 to A, 1 to B, and so on.