These are sufficient criteria to induce the undecidability of the halting problem of the machine class by a member of the class itself:
Exists x y in S such that x != y.
Exists f : S → S such that x != f(x)
Exists “loop forever”
Exists “halt”
Closed under function composition
Closed under conditional branching


Assume H is a machine in the class which takes as input a program p and input i and decides whether p halts on input i:

H(p,i) = 1 if program p halts on input i
 	= 0 otherwise

Since the class has a loop-forever instruction, a halt instruction, is closed under conditional branching and function composition, and H is assumed to be a machine in the class, then this diagonalization machine is also a machine in the class:
K(i) = if H(i,i) then loop forever, else halt

since it’s closed under function composition, we could define K1(i) = H(i,i)
since it’s closed under conditional branching, we could define K2(i) = if H(i,i) then A else B
since it has a loop-forever instruction and a halting instruction, then we can set A = loop-forever, B = halt: K3(i) = if H(i,i) then loop-forever else halt

Now let’s see our diagonalization in action:
H(K,K) = 1 if program K halts on input K, else 0
K(K) = if H(K,K) then loop forever, else halt

assume H(K,K) = 1, then K(K) loops forever, and so H(K,K) = 0, contradiction
assume H(K,K) = 0, then K(K) halts, and so H(K,K) = 1, contradiction

Paradox.
No class of machines that satisfies the constraints given above can solve its own halting problem.