On the methods of mechanical non-theorems (latest version)

Proof: Use a and b from the lemma above. Let l = b and m = a · b − b and k = a · b. Note that
from the conditions that 0 < a and 0 < b we get 0 < k. Moreover:
f k (α) = f a·b (α) by definition of k,
= f a·b−b+b (α) = f m+b (α) by basic arithmetic,
= f m+b (α) = f a·l+m+b (α) by the lemma,
= f a·b+(a·b−b)+b (α) by definition of m and n,
= f a·b+a·b (α) = f 2·k by basic arithmetic and the definition of k.
2.1.6 Vector-spaces over finite fields
The definition of a vector-space over a finite field is a ready made, finite and first-order definition
of the notion of a set of tuples with entries from a finite set. When the vector-space has dimension
m ∈ N, it is necessarily of the same form as the set of m-tuples with entries from the underlying
finite field. These vector-spaces can accordingly be assembled to n-tapes with entries from the
underlying field, for a chosen n.
A one-sorted definition of vector space over the field of integers modulo two follows.
Definition A vector space V over the minimal field is a tuple (V, +, −, 0, 0, 1) where V is a set, +
is a binary operation, −, 0, 1 are unary operations and where 0 is a constant. Moreover V is
1. an abelian group, i.e.
V |= ∀xyz[(x + y) + z = x + (y + z)],
V |= ∀x[x + 0 = x],
V |= ∀x[x + −(x) = 0],
V |= ∀xy[x + y = y + x],
2. left distributive, i.e.
V |= ∀xy[0(x + y) = 0(x) + 0(y)],
V |= ∀xy[1(x + y) = 1(x) + 1(y)],
3. right distributive, i.e.
V |= ∀x[0(x) = 0(x) + 0(x)],
V |= ∀x[0(x) = 1(x) + 1(x)],
V |= ∀x[1(x) = 0(x) + 1(x)],
V |= ∀x[1(x) = 1(x) + 0(x)],
qed
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