(f(x) − f(y)) − (f(x) − y)
= (ff(x) − f(y)) − (f(x) − y) (by Theorem 3.9)
= (ff(x) − (f(x) − y)) − f(y) (by (S3))
= f(f(x) − (f(x) − y)) − f(y) (by Definition 3.1)
= f(y − (y − f(x))) − f(y) (by (S2))
≤ f(y) − f(y) (by p(3), Theorem 3.9 and p(9))
= 0
It follows that (f(x)−f(y))−(f(x)−y) = 0 and f(x)−f(y) ≤ f(x)−y = f(x−y). Hence
f(x) − f(y) = f(x − y).
Let X be a subtraction algebra and f1, f2 two self-maps. We define f1 ◦ f2 : X → X by
(f1 ◦ f2)(x) = f1(f2(x))
for all x ∈ X.
Proposition 3.13. Let X be a subtraction algebra and f1, f2 two multipliers. Then f1 ◦ f2
is also a multiplier of X.
Proof. Let X be a subtraction algebra and f1, f2 two multipliers. Then we have
(f1 ◦ f2)(a − b) = f1(f2(a − b))
= (f1(f2(a) − b))
= f1(f2(a)) − b
= (f1 ◦ f2)(a) − b.
This completes the proof.
Let X be a subtraction algebra and f1, f2 two self-maps. We define (f1 ∧ f2)(x) by
(f1 ∧ f2)(x) = f1(x) ∧ f2(x)
for all x ∈ X.