Alternatively using z3 [0]:

  (declare-const a1 Bool)
  (declare-const a2 Bool)
  (declare-const a3 Bool)
  (declare-const a4 Bool)
  (declare-const a5 Bool)
  (declare-const a6 Bool)

  (assert (= a1 (and a2 (and a3 (and a4 (and a5 a6))))))
  (assert (= a2 (not (or a3 (or a4 (or a5 a6))))))
  (assert (= a3 (and a1 a2)))
  (assert (= a4 (or a1 (or a2 a3))))
  (assert (= a5 (not (or a1 (or a2 (or a3 a4))))))
  (assert (= a6 (not (or a1 (or a2 (or a3 (or a4 a5)))))))
  (assert (or a1 (or a2 (or a3 (or a4 (or a5 a6))))))

  (check-sat)
  (get-model)
  (exit)

The model is satisfiable where only a5 is true, anyone can run it themselves and play with it [1].

Additionally forcing a5 to be false by adding the following to the list of assertions:

  (assert (not a5))

And the model becomes unsatisfiable.

Edit: Finally, to make sure there's not some possible solution where a5 is true as well as another assertion you could alternatively add this assertion:

  (assert (and a5 (or a1 (or a2 (or a3 (or a4 (or a6)))))))

And the model also becomes unsatisfiable. So 5 only seems to be the correct answer.

[0] https://github.com/Z3Prover/z3

[1] https://rise4fun.com/Z3/1DMW

I posted a solution using the python z3 library, which might be easier to use for people used to python, here: https://news.ycombinator.com/item?id=21545143

That solution was not correct, see https://news.ycombinator.com/item?id=21568151

Using no libraries!

    a1([f,f,_,_,_,_]).
    a1([f,t,f,_,_,_]).
    a1([f,t,t,f,_,_]).
    a1([f,t,t,t,f,_]).
    a1([f,t,t,t,t,f]).
    a1([t,t,t,t,t,t]).

    a2([_,f,t,_,_,_]).
    a2([_,f,f,t,_,_]).
    a2([_,f,f,f,t,_]).
    a2([_,f,f,f,f,t]).
    a2([_,t,f,f,f,f]).

    a3([f,_,f,_,_,_]).
    a3([t,f,f,_,_,_]).
    a3([t,t,t,_,_,_]).

    a4([t,_,_,t,_,_]).
    a4([f,t,_,t,_,_]).
    a4([f,f,t,t,_,_]).
    a4([f,f,f,f,_,_]).

    a5([t,_,_,_,f,_]).
    a5([f,t,_,_,f,_]).
    a5([f,f,t,_,f,_]).
    a5([f,f,f,t,f,_]).
    a5([f,f,f,f,t,_]).

    a6([t,_,_,_,_,f]).
    a6([f,t,_,_,_,f]).
    a6([f,f,t,_,_,f]).
    a6([f,f,f,t,_,f]).
    a6([f,f,f,f,t,f]).
    a6([f,f,f,f,f,t]).

    solution(X) :- a1(X), a2(X), a3(X), a4(X), a5(X), a6(X).

Imho this would be a nice answer for stack exchange as well.

The fact that it is consistent does not mean that it is correct...