(c) Therefore we only have to show that for any s > 0

    If TM has not halted before s then D |^_ d_s

We shall prove this by induction on s.

(i)    Let s = 0.

        D contains d₀ by assumption. So D |^_ d₀

(ii)    Suppose the claim is true for some s, and TM has not halted before s+1.

        Then TM has not halted before s and won't halt at s.

        So by hypothesis D |^_ d_s for some description d_s

        We need to show that D |^_ d_s+1 for some description d_s+1

    .   Now A |= D Þ A |= d_s

        So at time s, TM is on p in state q_i scanning S_j.

    .   SinceTM doesn't halt at s there is an entry in its machine table of one of the three kinds:

        (a) (q_i, S_j, S_k, q_m)

        (b) (q_i, S_j, R, q_m)

        (c) (q_i, S_j, L, q_m)

        Consider each of these cases in turn

(a)    One of the sentences of D must be, by the construction of D,

        "txy [[Q_itx Ù S_jtx] ® [Q_mt'x Ù S_kt'x Ù (y¹x ® (S₀ty ® S₀t'y Ù ... Ù S_rty ® S_rt'y))]]

        Combine this with d_s, [Succ], and [Diff] to prove

        Q_me^s+1e^p Ù S_j1e^s+1e^p1 Ù ... Ù S_ke^s+1e^p Ù ... Ù S_jve^s+1e^pv

                        Ù "y[[y¹e^p1 Ù ... Ù y¹e^p Ù ... Ù y¹e^pv] ® S₀e^s+1y]

        which is an example of  d_s+1.