5.    We claim that D |^_ H iff this machine halts when given n as an input.

(Ü) The easy part.

A |= D by construction and the intention of the interpretation so D |^_ H Þ A |= H, but A |= H only if the machine halts with input n.

(Þ) The hard part.

(a). Introduce a type of sentence called a Description of Time t (labelled d_t) that tells you the status of the TM and its tape at that time.

A sentence of L is d_t if it looks like:

Q_ie^te^p Ù S_j1e^te^p1 Ù ... Ù S_je^te^p Ù ... Ù S_jve^te^pv Ù "y[[y¹e^p1 Ù ... Ù y¹e^p Ù ... Ù y¹e^pv] ® S₀e^ty]

with p₁, ..., p_v increasing indexes.

This is supposed to be understood as saying (in A) 'At time t the TM is on p in state q_i, and between cells p₁ and p_v there is some mix of symbols, and outside that range there are only zeroes.'

(b) Suppose TM(n) eventually halts.

Then there are t, i, p, j, such that at t the TM is on p in state q_i scanning S_j, and there is no quadruple beginning (q_i, S_j, ...) in TM's machine table.

Suppose D |^_ d_s a description of time s.

Now A |= D Þ A |= d_s

Two conjuncts of d_s are Q_ie^se^p and S_je^se^p, so

d_s |^_ $tx[Q_itx Ù S_jtx]

This is one of the disjuncts of H; so D |^_ H.