Applications: Mathematical Proofs and Automated Reasoning 237 Whenever we refer to a composition gf of two mappings f and g , we will assume that rng(f ) ⊆ dom(g ) Proposition 193 Composition of mappings is associative, that is, f (gh) = (f g )h, whenever either of these is defined Proposition 194 Let f : A → B , g : B → C be any mappings Then the following hold (a) (b) (c) (d) (e) If f and g are injective then gf is also injective If f and g are surjective then gf is also surjective In particular, the composition of bijections is a bijection If gf is injective then f is injective If gf is surjective then g is surjective Proposition 195 If f : A → B , g : B → C are bijective mappings then gf has an inverse, such that (gf )−1 = f −1 g −1 Proposition 196 A mapping f is injective iff for every two mappings g1 and g2 with dom(g1 ) = dom(g2 ) and cod(g1 ) = dom(f ) = cod(g2 ), the following left cancellation property holds: if f g1 = f g2 then g1 = g2 A mapping f is surjective iff for every two mappings g1 and g2 with dom(g1 ) = dom(g2 ) = cod(f ), the following right cancellation property holds: if g1 f = g2 f then g1 = g2 5.2.4 Binary relations and operations on them We have already discussed relations (predicates) of any number of arguments Being subsets of a given universal set, sets can be regarded as unary relations Here we focus on binary relations and operations on them First, let us summarize the basic terminology Given sets A and B , a binary relation between A and B is any subset of A × B In particular, a binary relation on a set A is any subset of A2 = A × A Given a binary relation R ⊆ A × B , if (a, b) ∈ R, we sometimes also write aRb and say that a is R-related to b Given sets A and B and a binary relation R A ì B : ã the domain of R is the set dom(R) = {a ∈ A | ∃b ∈ B (aRb)}; and • the range of R is the set rng(R) = {b ∈ B | ∃a ∈ A(aRb)} More generally, given subsets X ⊆ A and Y ⊆ B , we define • the image of X under R: R[X ] = {b ∈ B | ∃x ∈ X (xRb)}; and • the inverse image of Y under R: R−1 [Y ] = {a ∈ A | ∃y ∈ Y (aRy )} Notice that dom(R) = R−1 [B ] and rng(R) = R[A]