Computability and Logic

11.1. LOGIC AND TURING MACHINES 131
a, and M e ap will be either a conjunct of the description also (if e = 1) or a logical
implication of the description and Ɣ (if e = 0). Hence (14) and therefore D will be a
logical implication of Ɣ together with the description of time a. What if the machine
does not halt at time b = a + 1?
11.1 Lemma. If a ≥ 0, and b = a +1 is a time at which the machine has not (yet)
halted, then Ɣ together with the description of time a implies the description of time b.
Proof: The proof is slightly different for each of the four types of instructions
(print a blank, print a stroke, move left, move right). We do the case of printing a
stroke, and leave the other cases to the reader. Actually, this case subdivides into
the unusual case where there is already a stroke on the scanned square, so that the
instruction is just to change state, and the more usual case where the scanned square
is blank. We consider only the latter subcase.
So the description of time a looks like this:
(16)
Q i a &@ap & Maq 1 & Maq 2 & ...& Maq m &
∀x((x ≠ q 1 & x ≠ q 2 & ... & x ≠ q m ) →∼Max)
where p ≠ q r for any r,soƔ implies p ≠ q r by (9), and, by the argument given earlier,
Ɣ and (16) together imply ∼Map. The sentence in Ɣ corresponding to the applicable
instruction looks like this:
(17)
∀t∀x((Q i t &@tx & ∼Mtx) →
∃u(Stu &@ux & Mux & Q j u & ∀y((y ≠ x & Mty) → Muy)
& ∀y((y ≠ x & ∼Mty) →∼Muy))).
The description of time b looks like this:
(18)
Q j b &@bp & Mbp & Mbq 1 & Mbq 2 & ... & Mbq m &
∀x((x ≠ p & x ≠ q 1 & x ≠ q 2 & ...& x ≠ q m ) →∼Mbx).
And, we submit, (18) is a consequence of (16), (17), and Ɣ.
[Briefly put, the reason is this. Putting a for t and p for x in (17), we get
(Q i a &@ap & ∼Map) →
∃u(Sau &@up & Mup & Q j u &
∀y((y ≠ p & May) → Muy)&∀y((y ≠ p & ∼May) →∼Muy)).
Since (16) and Ɣ imply Q i a & @ap & ∼Map,weget
∃u(Sau &@up & Mup & Q j u &
∀y((y ≠ p & May) → Muy)&∀y((y ≠ p & ∼May) →∼Muy)).
By (10), Sau gives u = b, where b = a + 1, and we get
@bp & Mbp & Q j b &
∀y((y ≠ p & May) → Mby) &∀y((y ≠ p & ∼May) →∼Mby).
The first three conjuncts of this last are the same, except for order, as the first three
conjuncts of (18). The fourth conjunct, together with p ≠ q k from (9) and the conjunct