1-9:
    R(S, ...) always maps 0 to something ≥1.
    So M(R(S, ...) can't ever eval to 0.
    Thus M(M(R(S, ...) can never terminate.

10-17:
    R(C(S, ...) also always maps 0 to something ≥1.
    So M(M(R(C(S, ...) can't terminate.

18: M⁰(C¹(M¹(R²(P¹(1), P³(1))), S¹))
    Let f = R²(P¹(1), P³(1)), so that x18 ≡ M⁰(C¹(M¹(f), S¹)).
    f(0,b) = b.  f(a>0,b) = a-1.
    Let g = C¹(M¹(f), S¹) so that x18 ≡ M⁰(g).
    g = M¹(f)∘S¹, so g(a) = M¹(f)(a+1).
    Iterating to calculate g(a):
        i=0:    f(0,a+1) = a+1 ≠ 0
        i=1:    f(1,a+1) = 0
    So g(a) = 1.
    And so M⁰(g) ≡ M⁰(N¹(1)) and doesn't terminate.

19: M⁰(C¹(R²(S¹, Z³), Z¹, Z¹))
22: M⁰(C¹(R²(S¹, P³(1)), Z¹, Z¹))
25: M⁰(C¹(R²(S¹, P³(3)), Z¹, Z¹))
        These three are all M⁰(C¹(R²(S¹, any³), Z¹, Z¹)).
    Let f = R²(S¹, any³), so x## ≡ M⁰(C¹(f, Z¹, Z¹)).
    C¹(f, Z¹, Z¹) = λa: f(0, 0) = λa: 1 = N¹(1).
    So this is another M⁰(N¹(1)).

20: M⁰(C¹(R²(S¹, Z³), Z¹, S¹))
23: M⁰(C¹(R²(S¹, P³(1)), Z¹, S¹))
26: M⁰(C¹(R²(S¹, P³(3)), Z¹, S¹))
        These three are all M⁰(C¹(R²(S¹, any³), Z¹, S¹)).
    Let f=C¹(R²(S¹, any³), Z¹, S¹)
    f(a) = R²(S¹, any³)(0, a+1) = a+2.
    And M⁰(f) = ∞

21: M⁰(C¹(R²(S¹, Z³), Z¹, P¹(1)))
24: M⁰(C¹(R²(S¹, P³(1)), Z¹, P¹(1)))
27: M⁰(C¹(R²(S¹, P³(3)), Z¹, P¹(1)))
        These three are all M⁰(C¹(R²(S¹, any³), Z¹, P¹(1))).
    Let f=C¹(R²(S¹, any³), Z¹, P¹(1))
    f(a) = R²(S¹, any³)(0, a) = a+1.
    So this is M⁰(S¹) = ∞

28: M⁰(C¹(R²(S¹, P³(3)), P¹(1), S¹))
    Let f=C¹(R²(S¹, P³(3)), P¹(1), S¹).
    f(a) = R²(S¹, P³(3))(a, a+1).
    f(a=0) = a+2 = 2;  f(a>0) = a+1
    So f(a) can never be zero, and M(f) = ∞

29: M⁰(C¹(R²(S¹, P³(3)), P¹(1), P¹(1)))
    Let f=C¹(R²(S¹, P³(3)), P¹(1), P¹(1)).
    f(a) = R²(S¹, P³(3))(a, a).
    f(a=0) = a+1 = 1;  f(a>0) = a
    Once again, f(a) can never be zero.

30: M⁰(C¹(R²(P¹(1), Z³), Z¹, S¹))
31: M⁰(C¹(R²(P¹(1), P³(1)), Z¹, S¹))
        These two are both M⁰(C¹(R²(P¹(1), any³), Z¹, S¹))
    Let f=C¹(R²(P¹(1), any³), Z¹, S¹).
    f(a) = R²(P¹(1), any³)(0, a+1) = a+1
    So this is M⁰(S¹) = ∞